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Section 5.4 Indeterminate Form & L'H么pital's Rule

Subsection 5.4.1 Indeterminate Forms

Before we embark on introducing one more limit rule, we need to recall a concept from algebra. In your work with functions (see 颁丑补辫迟别谤听2) and limits (see 颁丑补辫迟别谤听4) we sometimes encountered expressions that were undefined, because they either lead to a contradiction or to numbers that are not in the set of numbers we started out with. Let us look at an example for either scenario to investigate the concept 鈥渦ndefined鈥� more deeply.

Example 5.30. Undefined because it Leads to Contradiction.

Suppose that

\begin{equation*} f(x)=\frac{1}{x}\text{.} \end{equation*}

What happens when \(x=0\text{?}\) Then \(f(0)=1/0\text{,}\) but \(1/0\) is undefined. Why is that? Let's assume this value is defined. This means that \(1/0\) is equal to some number, call it \(n\text{.}\) Then

\begin{equation*} \begin{split} \frac{1}{0} \amp = n \\ 1 \div 0 \amp = n \\ 1 \amp = n \times 0 \\ 1 \amp = 0 \end{split} \end{equation*}

Clearly, 1 is not equal to 0, and so this statement is a contradiction. In fact, if we analyze the satament

\begin{equation*} 1 = n \times 0\text{,} \end{equation*}

we notice that there is no number for \(n\) that will satisfy this equation. Therefore, \(1/0\) could not have been a number, and hence we say \(1/0\) is undefined. This is the reason why we write that the domain of \(f\) is given by

\begin{equation*} \mathcal{D}_{f} =\left\{ x \in \mathbb{R}\big\rvert x \neq 0 \right\}\text{.} \end{equation*}
Example 5.31. Different Number Set.

Suppose that \(f(x)=\sqrt{x-1}\) and that we are working over the real numbers. What happens when \(x=0\text{?}\) Then

\begin{equation*} f(0) = \sqrt{-1}\text{,} \end{equation*}

but \(\sqrt{-1}\) is undefined over the real numbers. Why is that? Let's assume this value is defined. Then by the definition of square root, there is a real number \(n\) such that \(-1 = n^2\text{.}\) Clearly, the square of a real number cannot produce a negative real number because positive 脳 positive and negative 脳 negative are both positive real numbers. In fact, \(\sqrt{-1}\) is the imaginary number \(i\text{,}\) which belongs to the set of complex numbers.

When we work out limit problems algebraically, we will often get as an initial answer something that is undefined. This is because the places where a function is undefined are the 鈥渋nteresting鈥� places to look for limits. For example, if

\begin{equation*} g(x)=\frac{x^{2}-9}{x-3}\text{,} \end{equation*}

then

\begin{equation*} g(3) = \frac{3^2-9}{3-3} = \frac{0}{0}\text{,} \end{equation*}

but

\begin{equation*} \lim_{x\to 3} g(x) = \lim_{x\to 3} \frac{x^2-9}{x-3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3} \left(x+3\right) = 6\text{.} \end{equation*}

The function \(g\) is a line with a hole at \(x=3\) and the limit showed us that this hole can be removed with the \(y\)-value 6 at \(x=3\) (see 贵颈驳听5.11).

Figure 5.11. The function \(g(x)=\frac{x^{2}-9}{x-3}\) is undefined at \(x=3\text{.}\)

However, we must remember that when we are calculating the limit of \(f(x)\) as \(x \to a\) we are not interested in the behavior of \(f(x)\) at \(a\text{,}\) but we want to know the behavior of \(f(x)\) around \(a\text{.}\) It is therefore important for us to identify an undefined value a of a function, and furthermore, to investigate whether the type of undefined value can tell us something about the behavior of the function around \(a\text{.}\)

Before we continue, we need to draw attention to a notation that we have been using when calculating limits. When we write \(f(x) \to 0\) as \(x \to a\text{,}\) we actually mean that \(f(x)\) gets arbitrarily close to zero as \(x\) gets closer and closer to \(a\text{.}\) However, the function value never reaches zero. Similarly, when we write \(f(x) \to \infty\) as \(x \to a\text{,}\) we actually mean that \(f(x)\) grows ever larger, without bound as \(x\) gets closer and closer to \(a\text{.}\) However, the function value never reaches infinity, since infinity is not even a number.

Limit Behaviour.

When calculating limits,

  1. 0 represents a number arbitrarily close to zero;

  2. \(+\infty\) represents an arbitrarily large positive number; and

  3. \(-\infty\) represents an arbitrarily large negative number.

Therefore, \(f(x) \to \frac{0}{0}\) as \(x \to a\) means that \(f(x)\) is a fraction for which both the numerator and the denominator get arbitrarily close to zero as \(x\) gets closer and closer to \(a\text{,}\) and \(f(x) \to \frac{\infty}{\infty}\) as \(x \to a\) means that \(f(x)\) is a fraction for which both the numerator and the denominator grow ever larger, without bound as \(x\) gets closer and closer to \(a\text{.}\) We also know from experience that some limits that demonstrate \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) behaviour work out to be real numbers, i.e. the limit exists, while others do not, as the following four examples remind us:

Example 5.32. Limit exists when 0/0.
\begin{equation*} \lim_{x\to 3} \frac{x^{2}-9}{x-3} \stackrel{\frac{0}{0}}{=} \lim_{x\to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x\to 3} (x+3) = 6 \end{equation*}
Example 5.33. Limit does not exist when 0/0.
\begin{equation*} \begin{split} \lim_{x\to 0^{+}} \frac{\sqrt{x+1}-1}{x^{2}} \stackrel{\frac{0}{0}}{=} \amp \lim_{x\to 0^{+}} \frac{\sqrt{x+1}-1}{x^{2}} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} \\ \amp = \lim_{x\to 0^{+}} \frac{1}{x\left(\sqrt{x+1}+1\right)} \stackrel{\frac{1}{0^{+}}}{=} \infty \end{split} \end{equation*}
Example 5.34. Limit exists when \(\infty\)/\(\infty\).
\begin{equation*} \lim_{x\to \infty} \frac{1-x}{2x} \stackrel{\frac{-\infty}{\infty}}{=} \lim_{x\to \infty} \frac{\frac{1}{x}-1}{2} = -\frac{1}{2} \end{equation*}
Example 5.35. Limit does not exist when \(\infty\)/\(\infty\).
\begin{equation*} \lim_{x\to \infty} \frac{1-x^{2}}{2x} \stackrel{\frac{-\infty}{\infty}}{=} \lim_{x\to \infty} \frac{\frac{1}{x}-x}{2} = -\infty \end{equation*}

Upon closer inspection of the undefined expressions 0/0 and \(\infty\)/\(\infty\text{,}\) we should realize that both terms are based on the division operation and ask ourselves whether there are other undefined expressions that we may encounter when taking limits. We therefore investigate arithmetic (\(a+b\text{,}\)\(a-b\text{,}\)\(ab\text{,}\)\(a/b\)) and exponentiating (\(a^b\)) operations where \(a\) and \(b\) are values that approach 0, 1, some arbitrary number \(n \neq 0,1\) or \(\infty\text{.}\) We leave it up to the reader to perform an exhaustive listing of all combinations, and instead limit ourselves to the combinations that are of interest as shown in 罢补产濒别听5.12.

\begin{equation*} \begin{array}{cccccc} 0+0 \amp \infty+\infty \amp 0 \cdot \infty \amp n \cdot \infty \amp 0^0 \amp 0^{\infty} \\[1ex] 0-0 \amp \infty-\infty \amp \frac{0}{\infty} \amp \frac{n}{\infty} \amp 1^0 \amp 1^{\infty} \\[1ex] 0 \cdot 0 \amp \pm\infty \cdot \pm\infty \amp \frac{\infty}{0} \amp \frac{\infty}{n} \amp n^0 \amp n^{\infty} \\[1ex] \frac{0}{0} \amp \frac{\pm\infty}{\pm\infty} \amp \amp \amp \infty^0 \amp \infty^{\infty} \end{array} \end{equation*}
Table 5.12. Arithmetic and Exponentiating Combinations.

We now encourage the reader to investigate each one of the terms shown in 罢补产濒别听5.12 and decide whether the undefined expression resolves to give a single number value or infinity (determinate form), or whether this cannot be determined (indeterminate form), all the while keeping in mind our earlier discussion on limit behaviour around \(x=a\text{.}\) We formally define this new terminology before we explore some terms together.

Definition 5.36. Determinate and Indeterminate Forms.

An undefined expression involving some operation between two quantities is called a determinate form if it evaluates to a single number value or infinity.

An undefined expression involving some operation between two quantities is called an indeterminate form if it does not evaluate to a single number value or infinity.

We will inspect multiplication more closely. Consider \(0 \times 0\text{.}\) Clearly, a number that is getting arbitrarily close to zero that is multiplied by another number that is getting arbitrarily close to zero gets even closer to zero, i.e. \(0 \times 0 \to 0\text{.}\) Now consider \(\infty \times \infty\text{.}\) Here, multiplying two values that are growing large without bound simply means that their product grows large without bound, i.e. \(\infty \times \infty \to \infty\text{.}\) Similarly, \(\left(-\infty\right)\times\infty\) means that the magnitude of the product grows large without bound and that \(\left(-\infty\right)\times\infty \to -\infty\text{.}\) What about \(n\times\infty\text{,}\) when \(n \neq 0\text{?}\) Here we need to differentiate between negative and positive values of \(n\text{:}\) If \(n>0\text{,}\) then \(n\times \infty \to \infty\text{,}\) and if \(n\lt 0\text{,}\) then \(n\times\infty \to -\infty\text{.}\) So far, we have only encountered determinate forms involving multiplication. Lastly, consider \(0\times\infty\text{.}\) Here, we have a number that is getting arbitrarily close to zero being multiplied with a value that is growing large without bounds. This is like two ends of a rope being tugged and we do not know which side is going to win. Therefore, \(0\times\infty\) is an expression that cannot be determined.

We leave the remaining terms up to the reader to investigate and simply present the determinate and indeterminate forms of the expressions from 罢补产濒别听5.12 in 罢补产濒别听5.13.

\begin{equation*} \begin{array}{cc} \textbf{ Determinate Forms} \amp \textbf{ Indeterminate Forms}\\ \\[0.25em] 0+0 \amp \infty - \infty \\[1em] 0-0 \amp \dfrac{0}{0} \\[1em] 0\cdot 0 \amp \dfrac{\pm \infty}{\pm \infty} \\[1em] \pm \infty \cdot \pm \infty \amp 0 \cdot \infty \\[1em] \dfrac{0}{\infty},\dfrac{n}{\infty} \amp 0^0 \\[1em] \dfrac{\infty}{0},\dfrac{\infty}{n} \amp \infty^0\\[1em] n\cdot\infty \ \ {\scriptstyle n\neq 0} \amp 1^{\infty}\\[1em] 0^{\infty} \amp \\[1em] n^{\infty} \ \ {\scriptstyle n \neq 1} \amp \\[1em] \infty^{\infty} \amp \end{array} \end{equation*}
Table 5.13. Determinate and Indeterminate Forms.

Subsection 5.4.2 L'H么pital's Rule for Finding Limits

We are now in a position to introduce one more technique for trying to evaluate a limit.

Definition 5.37. Limits of the Indeterminate Forms \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).

A limit of a quotient \(\lim\limits_{x\rightarrow a}\frac{f\left( x\right) }{g\left( x\right) }\) is said to be an indeterminate form of the type\(\frac{0}{0}\) if both \(f\left( x\right) \rightarrow 0\) and \(g\left( x\right) \rightarrow 0\) as \(x\rightarrow a\text{.}\) Likewise, it is said to be an indeterminate form of the type \(\frac{\infty }{\infty }\) if both \(f\left( x\right) \rightarrow \pm \infty\) and \(g\left( x\right) \rightarrow \pm \infty\) as \(x\rightarrow a\) (Here, the two \(\pm\) signs are independent of each other).

This theorem is somewhat difficult to prove, in part because it incorporates so many different possibilities, so we will not prove it here.

Note:

  1. There may be instances where we would need to apply L'H么pital's Rule multiple times, but we must confirm that \(\lim\limits_{x\to a}\dfrac{f'(x)}{g'(x)}\) is still indeterminate before we attempt to apply L'H么pital's Rule again.

  2. L'H么pital's Rule is also valid for one-sided limits and limits at infinity.

Notation when Applying L'H么pital's Rule.

We use the symbol \(\Heq\) to denote we are using l'H么pital's Rule in that step.

Example 5.39. L'H么pital's Rule and Indeterminate Form 0/0.

Compute \(\ds\lim_{x\to \pi}\frac{x^2-\pi^2}{\sin x}\text{.}\)

Solution

We use L'H么pital's Rule: Since the numerator and denominator both approach zero,

\begin{equation*} \lim_{x\to \pi}\frac{x^2-\pi^2}{\sin x}\Heq \lim_{x\to \pi}\frac{2x}{\cos x}\text{,} \end{equation*}

provided the latter exists. But in fact this is an easy limit, since the denominator now approaches \(-1\text{,}\) so

\begin{equation*} \lim_{x\to \pi}\frac{x^2-\pi^2}{\sin x}=\frac{2\pi}{-1} = -2\pi\text{.} \end{equation*}
Example 5.40. L'H么pital's Rule and Indeterminate Form \(\infty\)/\(\infty\).

Compute \(\ds\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}\text{.}\)

Solution

As \(x\) goes to infinity, both the numerator and denominator go to infinity, so we may apply L'H么pital's Rule:

\begin{equation*} \lim_{x\to \infty}\frac{2x^2-3x+7}{x^2+47x+1}\Heq \lim_{x\to \infty}\frac{4x-3}{2x+47}\text{.} \end{equation*}

In the second quotient, it is still the case that the numerator and denominator both go to infinity, so we are allowed to use L'H么pital's Rule again:

\begin{equation*} \lim_{x\to \infty}\frac{4x-3}{2x+47}\Heq\lim_{x\to \infty}\frac{4}{2}=2\text{.} \end{equation*}

So the original limit is 2 as well.

Example 5.41. L'H么pital's Rule and Indeterminate Form 0/0.

Compute \(\ds\lim_{x\to 0}\frac{\sec x - 1}{\sin x}\text{.}\)

Solution

Both the numerator and denominator approach zero, so applying L'H么pital's Rule:

\begin{equation*} \lim_{x\to 0}\frac{\sec x - 1}{\sin x}\Heq \lim_{x\to 0}\frac{\sec x\tan x}{\cos x}=\frac{1\cdot 0}{1}=0\text{.} \end{equation*}
Example 5.42. L'H么pital's Rule and Indeterminate Form \(\infty\)/\(\infty\).

Compute \(\ds\lim_{x\to 0^{+}} \dfrac{\frac{1}{x^2}}{\ln x}\text{.}\)

Solution

As \(x\) aproaches zero from the right, the numerator approaches \(+\infty\) and the denominator approaches \(-\infty\text{.}\) We may therefore apply L'H么pital's Rule:

\begin{equation*} \begin{split} \lim_{x\to 0^{+}} \dfrac{\frac{1}{x^2}}{\ln x} \amp \Heq \lim_{x \to 0^{+}} \frac{\frac{-2}{x^3}}{\frac{1}{x}} \\ \amp \Heq \lim_{x\to 0^{+}} \frac{-2}{x^{2}} \\ \amp = -\infty \end{split} \end{equation*}

Note: In order to decide which of two functions \(f\) and \(g\) grows faster as the independent variable, say \(x\text{,}\) becomes larger, we can apply the limit as \(x\) goes to infinity to the ratio \(f/g\) of these two functions. If the function \(f\) in the numerator grows faster, then the limit approaches infinity. If the function \(g\) in the denominator grows faster, then the limit approaches zero. If the functions have similar growth rates, then then the limit approaches a constant. This type of limit is readily computed using L'H么pital's Rule, and so L'H么pital's Rule is a useful tool to know.

We now exemplify this idea of growth rate. Let us have a closer look at the two functions \(f(x)=5x^3\) and \(g(x)=x^3\text{.}\) Then the function \(f(x)=5x^3\) grows exactly five times as fast as the function \(g(x)=x^3\text{.}\) However, the ratio of the two functions

\begin{equation*} \frac{f(x)}{g(x)} = \frac{5x^3}{x^3} = 5, x\neq 0\text{,} \end{equation*}

is a constant, and so both functions have fundamentally the same growth rate.

Subsection 5.4.3 Informally Extending L'H么pital's Rule

L'H么pital's Rule concerns limits of a quotient that are indeterminate forms. But not all functions are given in the form of a quotient. But all the same, nothing prevents us from re-writing a given function in the form of a quotient. Indeed, some functions whose given form involve either a product \(f\left( x\right) g\left( x\right)\) or a power \(f\left( x\right) ^{g\left( x\right) }\) carry indeterminacies such as \(0\cdot \left(\pm \infty\right)\) or \(1^{\pm \infty }\text{.}\) Something small times something numerically large (positive or negative) could be anything. It depends on how small and how large each piece turns out to be. A number close to 1 raised to a numerically large (positive or negative) power could be anything. It depends on how close to 1 the base is, whether the base is larger than or smaller than 1, and how large the exponent is (and its sign). We can use suitable algebraic manipulations to relate them to indeterminate quotients. We will illustrate with three examples, a product, a power and a difference.

Example 5.43. L'H么pital's Rule and Indeterminate Form \(0\times\infty\).

Compute \(\ds\lim_{x\to 0^+} x\ln x\text{.}\)

Solution

This doesn't appear to be suitable for L'H么pital's Rule, but it also is not 鈥渙bvious鈥�. As \(x\) approaches zero, \(\ln x\) goes to \(-\infty\text{,}\) so the product looks like:

\begin{equation*} (\hbox{something very small})\cdot(\hbox{something very large and negative})\text{.} \end{equation*}

This could be anything: it depends on how small and how large each piece of the function turns out to be. As defined earlier, this is a type of \(\pm0\cdot\infty\text{,}\) which is indeterminate. So we can in fact apply L'H么pital's Rule after re-writing it in the form \(\frac{\infty }{\infty }\text{:}\)

\begin{equation*} x\ln x = \frac{\ln x}{1/x}=\frac{\ln x}{x^{-1}}\text{.} \end{equation*}

Now as \(x\) approaches zero, both the numerator and denominator approach infinity (one \(-\infty\) and one \(+\infty\text{,}\) but only the size is important). Using L'H么pital's Rule:

\begin{equation*} \lim_{x\to 0^+} {\ln x\over x^{-1}}\Heq \lim_{x\to 0^+} {1/x\over -x^{-2}} =\lim_{x\to 0^+} {1\over x}(-x^2)= \lim_{x\to 0^+} -x = 0\text{.} \end{equation*}

One way to interpret this is that since \(\ds\lim_{x\to 0^+}x\ln x = 0\text{,}\) the \(x\) approaches zero much faster than the \(\ln x\) approaches \(-\infty\text{.}\)

Finally, we illustrate how a limit of the type \(1^\infty\) can be indeterminate.

Example 5.44. L'H么pital's Rule and Indeterminate Form \(1^{\infty}\).

Compute \(\ds{\lim_{x\to 1^+}x^{1/(x-1)}}\text{.}\)

Solution

Plugging in \(x=1\) (from the right) gives a limit of the type \(1^\infty\text{.}\) To deal with this type of limit we will use logarithms. Let

\begin{equation*} L=\lim_{x\to 1^+}x^{1/(x-1)}\text{.} \end{equation*}

Now, take the natural log of both sides:

\begin{equation*} \ln L=\lim_{x\to 1^+}\ln\left(x^{1/(x-1)}\right)\text{.} \end{equation*}

Using log properties we have:

\begin{equation*} \ln L=\lim_{x\to 1^+}\frac{\ln x}{x-1}\text{.} \end{equation*}

The right side limit is now of the type \(0/0\text{,}\) therefore, we can apply L'H么pital's Rule:

\begin{equation*} \ln L=\lim_{x\to 1^+}\frac{\ln x}{x-1}\Heq\lim_{x\to 1^+}\frac{1/x}{1}=1 \end{equation*}

Thus, \(\ln L=1\) and hence, our original limit (denoted by \(L\)) is: \(L=e^1=e\text{.}\) That is,

\begin{equation*} L=\lim_{x\to 1^+}x^{1/(x-1)}=e\text{.} \end{equation*}

In this case, even though our limit had a type of \(1^\infty\text{,}\) it actually had a value of \(e\text{.}\)

Example 5.45. L'H么pital's Rule and Indeterminate Form \(\infty-\infty\).

Compute \(\lim\limits_{x\to 0^{+}} \left(\dfrac{1}{\sin x} - \dfrac{1}{x}\right)\text{.}\)

Solution

As \(x\) approaches zero from the right,

\begin{equation*} \frac{1}{\sin x} - \frac{1}{x} \to \infty - \infty\text{.} \end{equation*}

This is not a form on which we know we can use L'H么pital's Rule, however, if we combine the fractions, the problem becomes

\begin{equation*} \lim_{x\to 0^{+}} \frac{x-\sin x}{x\sin x}\text{,} \end{equation*}

which gives us the indeterminate form \(0/0\text{.}\) We can now apply L'H么pital's Rule twice:

\begin{equation*} \begin{split} \lim_{x\to 0^{+}} \frac{x-\sin x}{x\sin x} \amp \Heq \lim_{x\to 0^{+}} \frac{1-\cos x}{\sin x + x\cos x} \\ \amp\Heq \lim_{x\to 0^{+}} \frac{\sin x}{2\cos x - x\sin x}\\ \amp= \frac{0}{2-0} = 0.\end{split} \end{equation*}
Exercises for Section聽5.4.
Compute the following limits.
  1. \(\ds\lim_{x\to 0} \frac{\cos x -1}{\sin x}\) Answer
    0
    Solution

    We see that the limit

    \begin{equation*} \lim_{x\to 0} \frac{\cos x -1}{\sin x} \end{equation*}

    is of the indeterminate form \(0/0\text{.}\) We can therefore apply L'H么pital's Rule once to find,

    \begin{equation*} \lim_{x\to 0} \frac{\cos x -1}{\sin x} \Heq \lim_{x\to 0}\frac{-\sin x}{\cos x} = \frac{0}{1} = 0\text{.} \end{equation*}
  2. \(\ds\lim_{x\to \infty} \frac{e^x}{x^3}\) Answer
    \(\infty\)
    Solution

    We now see the indeterminate form \(\infty/\infty\) and apply L'H么pital's Rule until the form becomes determinate.

    \begin{equation*} \lim_{x\to \infty} \frac{e^x}{x^3} \Heq \lim_{x \to \infty} \frac{e^x}{3x^2} \Heq \dots \Heq \lim_{x\to \infty} \frac{e^x}{6} = \infty\text{.} \end{equation*}
  3. \(\ds\lim_{x\to \infty} \frac{\ln x}{x}\) Answer
    0
    Solution

    The limit is of the indeterminate form \(\infty/\infty\text{,}\) and so we apply L'H么pital's Rule until the form becomes determinate.

    \begin{equation*} \lim_{x\to \infty} \frac{\ln x}{x} \Heq \lim_{x\to\infty} \frac{1/x}{\frac{1}{2}x^{-1/2}} = \lim_{x\to\infty} \frac{2}{\sqrt{x}} = 0\text{.} \end{equation*}
  4. \(\ds\lim\limits_{x\to\infty} \dfrac{\ln x}{x} \) Answer
    0
    Solution

    To evaluate \(\lim\limits_{x\to\infty} \dfrac{\ln x}{x}\text{,}\) we see that this is of the indeterminate form \(\dfrac{\infty}{\infty}\) and so we can apply L'H么pital's Rule:

    \begin{equation*} \lim\limits_{x\to\infty} \dfrac{\ln x}{x} \Heq \lim_{x\to\infty} \frac{\diff{}{x}\ln x}{\diff{}{x} x}= \lim_{x\to\infty} \frac{\frac{1}{x}}{1} = \lim_{x\to\infty} \frac{1}{x} = 0\text{.} \end{equation*}
  5. \(\ds\lim\limits_{x\to 0} \frac{\sqrt{9+x}-3}{x} \) Answer
    1/6
    Solution

    The limit is of the indeterminate form \(0/0\) and so we apply L'H么pital's Rule:

    \begin{equation*} \lim\limits_{x\to 0} \frac{\sqrt{9+x}-3}{x} \Heq \lim\limits{x\to 0} \frac{\frac{1}{2\sqrt{9+x}}}{1} = \frac{1}{2(3)} = \frac{1}{6}\text{.} \end{equation*}
  6. \(\ds\lim_{x\to 2} \frac{2-\sqrt{x+2}}{4-x^2}\) Answer
    1/16
    Solution

    We see that the limit is of the indeterminate form \(0/0\text{,}\) and so we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 2} \frac{2-\sqrt{x+2}}{4-x^2} \Heq \lim_{x\to 2} \frac{-\frac{-1}{2\sqrt{x+2}}}{-2x} = \frac{1}{16}\text{.} \end{equation*}
  7. \(\ds\lim_{x\to 1} \frac{\sqrt{x} - 1}{\sqrt[3]{x}-1}\) Answer
    3/2
    Solution

    The limit is of the indeterminate form \(0/0\text{,}\) and so we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 1} \frac{\sqrt{x} - 1}{\sqrt[3]{x}-1} \Heq \lim_{x\to 1} \frac{\frac{x^{-1/2}}{2}}{\frac{x^{-2/3}}{3}} = \frac{3}{2}\text{.} \end{equation*}
  8. \(\ds\lim_{x\to 0} \frac{x^2}{\sqrt{2x+1}-1} \) Answer
    0
    Solution

    The limit is of the interderminate form \(0/0\text{,}\) and so we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 0} \frac{x^2}{\sqrt{2x+1}-1} \Heq \lim_{x\to 0} \frac{2x}{(2x+1)^{-1/2}} = 0\text{.} \end{equation*}
  9. \(\ds \lim_{u to 1} \frac{(u-1)^3}{(1/u)-u^2+(3/u)-3}\) Solution

    The limit is of the indeterminate form \(0/0\text{,}\) and so we apply L'H么pital's Rule:

    \begin{equation*} \lim_{u to 1} \frac{(u-1)^3}{(1/u)-u^2+(3/u)-3} \Heq \lim_{u \to 1} \frac{3(u-1)^2}{-u^{-2}-2u - 3u^{-2}} = 0\text{.} \end{equation*}
  10. \(\ds\lim_{x \to 0} \frac{2+(1/x)}{3-(2/x)}\) Answer
    -1/2
    Solution

    The limit is of the indeterminate form \(\infty/\infty\text{,}\) and so we could apply L'H么pital's Rule. However, notice that

    \begin{equation*} \lim_{x \to 0} \frac{2+(1/x)}{3-(2/x)} \Heq \lim_{x \to 0} \frac{2-x^{-2}}{3+2x^{-2}} = ..\text{.} \end{equation*}

    and so this procedure will not lead to a determinate form. Instead, let \(t= 1/x\text{.}\) Then

    \begin{equation*} \lim_{x\to 0^+} \frac{2+(1/x)}{3-(2/x)} = \lim_{t\to \infty} \frac{2+t}{3-2t} \Heq \lim_{t\to \infty}\frac{1}{-2} = -\frac{1}{2}\text{,} \end{equation*}

    and similarly,

    \begin{equation*} \lim_{x\to 0^-} \frac{2+(1/x)}{3-(2/x)} = \lim_{t\to -\infty} \frac{2+t}{3-2t} \Heq \lim_{t\to \infty}\frac{1}{-2} = -\frac{1}{2}\text{.} \end{equation*}

    Hence,

    \begin{equation*} \lim_{x\to 0} \frac{2+(1/x)}{3-(2/x)} = -\frac{1}{2}\text{.} \end{equation*}
    Answer
    5
  11. \(\ds\lim_{x\to 0^+} \frac{1+5/\sqrt{x}}{2+1/\sqrt{x}}\) Solution

    The limit is of the indeterminate form \(\infty/\infty\text{:}\)

    \begin{equation*} \lim_{x\to 0^+} \frac{1+5/\sqrt{x}}{2+1/\sqrt{x}} = \lim_{x\to 0^{+}} \frac{\sqrt{x} + 5}{2\sqrt{x}+1} = 5\text{.} \end{equation*}
  12. \(\ds \lim_{x\to \pi/2} \frac{\cos x}{(\pi/2)-x} \) Answer
    1
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to \pi/2} \frac{\cos x}{(\pi/2)-x} \Heq \lim_{x\to \pi/2} = \frac{-\sin x}{-1} = 1\text{.} \end{equation*}
  13. \(\ds\lim_{x\to 0} \frac{x^2}{e^x-x-1}\) Answer
    2
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule twice:

    \begin{equation*} \lim_{x\to 0} \frac{x^2}{e^x-x-1} \Heq \lim_{x\to 0} \frac{2x}{e^x-1} \Heq \lim_{x\to 0} \frac{2}{e^x} = 2\text{.} \end{equation*}
  14. \(\ds \lim_{x\to 1} \frac{\ln x}{x-1} \) Answer
    1
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 1} \frac{\ln x}{x-1} \Heq \lim_{x\to 1} \frac{1/x}{1} = 1\text{.} \end{equation*}
  15. \(\ds\lim_{x\to 0} \frac{\ln(x^2+1)}{x} \) Answer
    0
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 0} \frac{\ln(x^2+1)}{x} \Heq \frac{\frac{2x}{x^2+1}}{1} = 0\text{.} \end{equation*}
  16. \(\ds\lim_{x\to 1} \frac{x\ln x}{x^2-1}\) Answer
    1/2
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 1} \frac{x\ln x}{x^2-1} \Heq \lim_{x\to 1} \frac{\ln x + 1}{2x}= \frac{1}{2}\text{.} \end{equation*}
  17. \(\ds\lim_{x\to 0} \frac{\sin(2x)}{\ln(x+1)} \) Answer
    2
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 0} \frac{\sin(2x)}{\ln(x+1)} \Heq \frac{2\cos x}{1/(x+1)} = 2\text{.} \end{equation*}
  18. \(\ds\lim_{x\to 1} \frac{\sqrt{x}-1}{x-1} \) Answer
    1/2
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 1} \frac{\sqrt{x}-1}{x-1} \Heq \frac{\frac{1}{2\sqrt{x}}}{1} = \frac{1}{2}\text{.} \end{equation*}
  19. \(\ds \lim_{x\to 0} \frac{\sqrt{x+1}-1}{\sqrt{x+4}-2} \) Answer
    2
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 0} \frac{\sqrt{x+1}-1}{\sqrt{x+4}-2} \Heq \lim_{x\to 0} \frac{\sqrt{x+4}}{\sqrt{x+1}} = 2\text{.} \end{equation*}
  20. \(\ds \lim_{x\to 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x+1}-1} \) Answer
    0
    Solution

    The limit is of the indeterminate form \(0/0\text{.}\) Therefore, we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x+1}-1} \Heq \lim_{x\to 0} \frac{2x\sqrt{x+1}}{\sqrt{x^2+1}} = 0\text{.} \end{equation*}
  1. \(\ds \lim_{x\to 0^+} \sqrt{x}\ln x \) Answer
    0
    Solution

    The limit

    \begin{equation*} \lim_{x\to 0^+} \sqrt{x}\ln x \end{equation*}

    is of the form \(0 \cdot (-\infty)\text{,}\) therefore we need to transform the limit in order to use L'H么pital's Rule. Let \(t=1/x\text{.}\) Then,

    \begin{equation*} \lim_{x\to 0^+} \sqrt{x}\ln x = \lim_{t \to +\infty} \frac{\ln\left(\frac{1}{t}\right)}{\sqrt{t}} \end{equation*}

    which is of the form \(-\infty/\infty\text{.}\) We apply L'H么pital's Rule until the form becomes determinate.

    \begin{equation*} \begin{split} \lim_{t \to \infty} \frac{\ln\left(\frac{1}{t}\right)}{\sqrt{t}} \amp = \lim_{t \to \infty} \frac{\diff{}{t}\left(\ln(1)-\ln(t)\right)}{\frac{1}{2}t^{-1/2}} \\ \amp = \lim_{t \to \infty} \frac{-1/t}{\frac{1}{2}t^{-1/2}}\\ \amp = \lim_{t \to \infty} \frac{-2}{\sqrt{t}} = 0. \end{split} \end{equation*}
  2. \(\ds\lim_{x\to 0} \frac{(1-x)^{1/4} -1}{x}\) Answer
    -1/4
    Solution

    The limit is of the indeterminate form \(0/0\text{,}\) and so we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 0} \frac{(1-x)^{1/4} -1}{x} \Heq \lim_{x\to 0}\frac{-\frac{1}{4} (1-x)^{-3/4}}{1} = -\frac{1}{4}\text{.} \end{equation*}
  3. \(\ds \lim_{t\to 0} \left(t+\frac{1}{t}\right)\left((4-t)^{3/2} - 8\right)\) Answer
    -3
    Solution

    The limit is of the form \(\infty \cdot 0\text{,}\) and so we need to transform the limit in order to apply L'H么pital's Rule. Let

    \begin{equation*} \lim_{t\to 0} \left(t+\frac{1}{t}\right)\left((4-t)^{3/2} - 8\right) = \lim_{t\to 0} t\left((4-t)^{3/2} - 8\right) + \lim_{t\to 0} \frac{1}{t}\left((4-t)^{3/2} - 8\right)\text{.} \end{equation*}

    The first limit we can readily evaluate as

    \begin{equation*} \lim_{t\to 0} t\left((4-t)^{3/2} - 8\right) = 0\text{.} \end{equation*}

    For the second limit, we can now apply L'H么pital's Rule:

    \begin{equation*} \lim_{t\to 0} \frac{(4-t)^{3/2} - 8}{t} \Heq \lim_{t\to 0} \frac{-3/2 \sqrt{4-t}}{1} = -3\text{.} \end{equation*}
  4. \(\ds\lim_{t\to 0^+} \left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)\left(\sqrt{t+1}-1\right)\) Answer
    1/2
    Solution

    The limit is of the form \(\infty \cdot 0\text{,}\) and so we must transform the limit in order to apply L'H么pital's Rule. Let

    \begin{equation*} \lim_{t\to 0^+} \left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)\left(\sqrt{t+1}-1\right) = \lim_{t\to 0^+} \frac{\sqrt{t+1}-1}{t} + \lim_{t\to 0^+} \frac{\sqrt{t+1}-1}{\sqrt{t}}\text{.} \end{equation*}

    Both limits are of the form \(0/0\) and so we can apply L'H么pital's Rule to each:

    \begin{equation*} \lim_{t\to 0^+} \frac{\sqrt{t+1}-1}{t} \Heq \lim_{t\to 0^+} \frac{1}{2\sqrt{t+1}} = \frac{1}{2}\text{.} \end{equation*}

    We now apply L'H么pital's Rule to the second limit:

    \begin{equation*} \lim_{t\to 0^+} \frac{\sqrt{t+1}-1}{\sqrt{t}} \Heq \lim_{t\to 0^+} \frac{\sqrt{t}}{\sqrt{t+1}} = 0\text{.} \end{equation*}

    Therefore,

    \begin{equation*} \lim_{t\to 0^+} \left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)\left(\sqrt{t+1}-1\right) = \frac{1}{2}\text{.} \end{equation*}
  5. \(\ds \lim_{x\to 0} \frac{e^x-1}{x}\) Answer
    1
    Solution

    This limit is of the indeterminate form \(\dfrac{0}{0}\text{,}\) and so we apply L'H么pital's Rule:

    \begin{equation*} \lim_{x\to 0} \frac{e^x-1}{x} \Heq \lim_{x\to 0} \frac{\diff{}{x} \left(e^x-1\right)}{\diff{}{x} x} = \lim_{x\to 0} \frac{e^x}{1} = e^0 = 1\text{.} \end{equation*}
  6. \(\ds\lim\limits_{x\to 1} \dfrac{x^{1/4}-1}{x}\) Answer
    0
    Solution

    We evaluate:

    \begin{equation*} \lim\limits_{x\to 1} \dfrac{x^{1/4}-1}{x} = \dfrac{1-1}{1} = 0\text{.} \end{equation*}

    Notice that \(\dfrac{0}{1}\) is not an indeterminate form and we do not apply L'H么pital's Rule.

Discuss what happens if we try to use L'H么pital's Rule to find \(\ds \lim_{x\to \infty} \frac{x+\sin x}{x+1}\text{.}\) Solution

The limit

\begin{equation*} \lim_{x\to\infty} \frac{x+\sin x}{x+1} \end{equation*}

is of the indeterminate form \(\infty/\infty\text{.}\) And so we could try to apply L'H么pital's Rule:

\begin{equation*} \lim_{x\to\infty} \frac{x+\sin x}{x+1} \Heq \frac{1+\cos x}{1} = 1 +\lim_{x\to\infty} \cos x = DNE\text{.} \end{equation*}
\begin{equation*} \lim_{x\to\infty} \frac{x}{x+1} + \lim_{x\to \infty} \frac{\sin x}{x+1} = 1 + \lim_{x\to\infty}\text{.} \end{equation*}
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