Lesson Part 1

In This Module

• We will introduce a tool, called l'Hospital's rule, which is used to evaluate limits involving quotients of functions that cannot be evaluated using the limit laws.
• L'Hospital's rule uses derivatives to transform a limit with an indeterminate form (\(0\) over \(0\), or infinity over infinity) into a determinate form which, in many cases, can be easily evaluated.

Indeterminate Form

How do we sketch the graph of the function \(f(x)= \dfrac{\ln(x)}{x-1}\)?

Since the domain of \(\ln(x)\) is \(x\gt 0\), the function \(f(x)\) is defined for all positive real numbers, except for \(x= 1\), as the denominator is equal to \(0\) when \(x=1\).

To sketch the graph of the function, we need to analyze

• the behaviour of the function as \(x\) approaches the left endpoint (\(x=0\)) from the right;
• the discontinuity \(x=1\) from both the left and the right; and
• its behaviour as \(x\) approaches positive infinity.

This means examining the following four limits.

1. \(\displaystyle \lim_{x \rightarrow 0^{+}} \frac{\ln(x)}{x-1}\)

   As \(x\) approaches \(0\) from the right,

   • the function \(\ln (x)\) approaches \(-\infty\) and
   • the function \(x-1\) approaches \(-1\).

   To keep track of this information concerning the limits of the numerator and the denominator, we use the following informal notation \( \left(\dfrac{-\infty}{-1}\right)\). Here we have a fraction enclosed by brackets where the limit of the numerator is on top and the limit of the denominator is on the bottom.

2. \(\displaystyle \lim_{x \rightarrow 1^{-}} \frac{\ln(x)}{x-1}\)

   As \(x\) approaches \(1\) from the left,

   • the function \(\ln(x)\) approaches \(0\) since \(\ln(1)=0\) and
   • the function \(x-1\) approaches \(1-1\), which is \(0\).

   So again, we record this information using our fraction \(\left(\dfrac{0}{0}\right)\).

3. \(\displaystyle \lim_{x \rightarrow 1^{+}} \frac{\ln(x)}{x-1}\)

   As \(x\) approaches \(1\) from the right,

   • the function \(\ln(x)\) approaches \(0\) and
   • the function \(x-1\) approaches \(0\).

   This is similar to the previous limit, and it is also has the informal notation \(\left(\dfrac{0}{0}\right)\).

4. \(\displaystyle \lim_{x \rightarrow \infty} \frac{\ln(x)}{x-1}\)

   As \(x\) approaches \(\infty\),

   • the function \(\ln(x)\) approaches \(\infty\) and
   • the function \(x-1\) approaches \(\infty\).

   Again, we keep track of these limits using our informal notation \(\left(\dfrac{\infty}{\infty}\right)\).

If we look at the first limit \(\displaystyle \lim_{x \rightarrow 0^{+}} \frac{\ln(x)}{x-1}\), the numerator is approaching a very big negative number, and the denominator is approaching the constant \(-1\), and so the value of the first limit is \(\infty\). But it is not so clear how to evaluate the other limits.

We cannot use the limit law for quotients as the limits in the denominator are either \(0\) or \(\infty\).

It turns out that each of these limits exists (and the two limits involving \(x=1\) are equal), but the values are not at all obvious.

Indeterminate Form of Type \(\frac{0}{0}\)

In general, if we have a limit of the form

\[\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)}\]

where \( \displaystyle \lim_{x \rightarrow a} f(x) = 0\) and \(\displaystyle \lim_{x \rightarrow a}g(x) = 0\), then the limit of the quotient may or may not exist.

We call such a limit an indeterminate form of type \(\frac{0}{0}\).

Indeterminate Form of Type \(\frac{\infty}{\infty}\)

In general, if we have a limit of the form

\[\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)}\]

where \(\displaystyle \lim_{x \rightarrow a} f(x) \rightarrow \pm \infty\) and \(\displaystyle \lim_{x \rightarrow a}g(x) \rightarrow \pm \infty\), then the limit of the quotient may or may not exist.

We call such a limit an indeterminate form of type \(\frac{\infty}{\infty}\).

Note:

These definitions also apply to one-sided limits and limits at infinity and negative infinity.

In other words, the above statement also holds if “\(x \rightarrow a\)” is replaced with “\(x \rightarrow a^{+}\),” “\(x \rightarrow a^{-}\),” “\(x \rightarrow \infty\),” or “\(x \rightarrow -\infty\).”

L'Hospital's rule will provide a method to calculate the limit of quotients of this form, given that the limit indeed exists or approaches \(\pm \infty\).

So how might we deal with indeterminant limits in general? Here's an informal idea to motivate it.

Informal Idea: Evaluating \(\displaystyle \lim_{x\rightarrow a} \tfrac{f(x)}{g(x)}\)

Suppose that \(f(x)\) and \(g(x)\) both approach infinity as \(x\) approaches \(a\).

In other words, the limit of the quotient \(\dfrac{f(x)}{g(x)}\) is an indeterminate form of type \(\dfrac{\infty}{\infty}\).

If \(f(x)\) approaches infinity much “faster” than \(g(x)\), then we might expect the quotient \(\dfrac{f(x)}{g(x)}\) to approach infinity as well.

On the other hand, if \(g(x)\) approaches infinity much “faster” than \(f(x)\), then we might expect the quotient \(\dfrac{f(x)}{g(x)}\) to approach \(0\).

If \(f(x)\) and \(g(x)\) approach infinity at “similar” rates, then perhaps we expect the quotient to approach \(1\) or some non-zero constant.

In any case, if the function values themselves cannot give us a clue of the behaviour of the quotient, then it may be helpful to also compare the rates of change of \(f(x)\) and \(g(x)\).

This is the content of l'Hospital's rule.

Lesson Part 2

L'Hospital's Rule

L'Hospital's Rule

Suppose that \(f(x)\) and \(g(x)\) are differentiable and that \(g'(x) \neq 0\) near \(x=a\) (except for possibly at \(x=a\)). If the limit

\[\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)}\]

is an indeterminate form of type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then

\[\displaystyle \lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \displaystyle \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} \]

provided that the limit on the right side exists (or approaches \(\infty\) or \(-\infty\)).

L'Hopital's rule states that, under certain conditions, the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

Notes:

1. L'Hospital's rule is also valid for one-sided limits and limits at infinity or negative infinity.

   In other words, the above statement also holds if “\(x \rightarrow a\)” is replaced with “\(x \rightarrow a^{+}\),” “\(x \rightarrow a^{-}\),” “\(x \rightarrow \infty\),” or
   “\(x \rightarrow -\infty\).”

2. When applying l'Hospital's rule, we find the derivatives of the numerator, \(f(x)\), and the denominator, \(g(x)\), separately. We do not take the derivative of the quotient, \(\dfrac{f(x)}{g(x)}\).

Examples

Example 1

Use both the limit laws and l'Hospital's rule to evaluate the following limit:

\[\displaystyle \lim_{x \rightarrow \infty} \dfrac{3x^{2} - 2}{2x^{2} +1}\]

Solution

Method 1: Using Limit Laws

First, we divide the numerator and the denominator by the largest power of \(x\) appearing in the denominator and then simplify:

\[\begin{align*} \lim_{x \rightarrow \infty} \dfrac{3x^{2} - 2}{2x^{2} +1} & = \lim_{x \rightarrow \infty} \dfrac{\frac{3x^{2}}{x^{2}} - \frac{2}{x^{2}}}{\frac{2x^{2}}{x^{2}} + \frac{1}{x^{2}}} \\ & = \lim_{x \rightarrow \infty} \dfrac{3 - \frac{2}{x^{2}}}{2 + \frac{1}{x^{2}}} \\ & = \dfrac{3-0}{2+0} & \text{since } \lim_{x \rightarrow \infty} \frac{1}{x^{2}} = \lim_{x \rightarrow \infty} \frac{2}{x^{2}} = 0 \\ & = \dfrac{3}{2} \end{align*}\]

Method 2: Using l'Hospital's Rule

Let \(f(x) = 3x^{2} - 2\), the function in the numerator, and \(g(x) = 2x^{2}+1\), the function in the denominator.

Since \(f(x)\) and \(g(x)\) are both polynomials opening upward, \(\displaystyle \lim_{x \rightarrow \infty} f(x) \rightarrow \infty\) and \(\displaystyle \lim_{x \rightarrow \infty} g(x) \rightarrow \infty\), so we have an indeterminate form of type \(\frac{\infty}{\infty}\).

Since \(f'(x) = 6x\) and \(g'(x) = 4x\), we have that \(f\) and \(g\) are differentiable and that \(g'(x) = 0\) only when \(x=0\).

Since we're interested in the limit as \(x\) approaches infinity, \(g'(x)\) is non-zero for all large numbers, and so we can apply L'Hospital's rule.

By l'Hospital's rule, we have

\[\begin{align*} \lim_{x\rightarrow \infty} \dfrac{3x^{2} - 2}{2x^{2} +1} & = \lim_{x\rightarrow \infty} \dfrac{\dfrac{d}{dx}\Big(3x^{2} - 2\Big)}{\dfrac{d}{dx}\Big(2x^{2} +1\Big)} \\ & = \lim_{x\rightarrow \infty} \dfrac{6x}{4x} \end{align*}\]

provided that the limit on the right-hand side exists (or approaches \(\infty\) or \(-\infty\)).

This limit indeed exists and can be easily calculated:

\[\displaystyle \lim_{x\rightarrow \infty} \dfrac{6x}{4x} = \displaystyle \lim_{x\rightarrow \infty} \dfrac{6}{4} = \frac{3}{2}\]

As we would expect, both methods produce the same answer.

Lesson Part 3

Proof of a Special Case of l'Hospital's Rule

We will not prove the full version of l'Hospital's rule as stated, but we will prove that the result holds in the case where \(f(a) = g(a) = 0\), \(f'\) and \(g'\) are continuous, and \(g'(a)\neq 0\).

The more general proof is much more difficult.

Given these assumptions, let's attempt to evaluate \(\displaystyle \lim_{x \rightarrow a} \dfrac{f(x)}{g(x)}\).

Since \(f(a) = g(a) = 0\), this is an indeterminate form of type \(\dfrac{0}{0}\). Having \(f(a) = g(a) = 0\) also allows us to use an algebra trick here.

\[\begin{align*} \lim_{x \rightarrow a} \dfrac{f(x)}{g(x)} & = \lim_{x \rightarrow a} \dfrac{f(x)-0}{g(x)-0} & \text{because}\mp 0 \text{ does not change the values} \\ & = \lim_{x \rightarrow a} \dfrac{f(x)- f(a)}{g(x)-g(a)} & \text{because } f(a) = g(a) = 0 \\ & = \lim_{x \rightarrow a} \dfrac{\frac{f(x)- f(a)}{x-a}}{~\frac{g(x)-g(a)}{x-a}~} \end{align*}\]

We can divide the numerator and denominator by the same factor, \(x - a\), since we are considering the limit as \(x\to a\), thus \(x\neq a\) or \(x-a\neq 0\).

Now these limits should look very familiar. Provided that the limit as \(x\to a\) of the term in the numerator and the denominator exists, and the limit in the denominator is non-zero, we can use the quotient law for limits and rewrite the equation.

\[\begin{align*} \lim_{x \rightarrow a} \dfrac{\frac{f(x)- f(a)}{x-a}}{~\frac{g(x)-g(a)}{x-a}~} & = \dfrac{~\displaystyle \lim_{x \rightarrow a} \dfrac{f(x)- f(a)}{x-a}~}{~\displaystyle \lim_{x \rightarrow a} \dfrac{g(x)-g(a)}{x-a}~} \\[1mm] & = \dfrac{f'(a)}{g'(a)} \end{align*}\]

We recognize the numerator as the derivative of \(f\) at \(a\). And likewise, the limit in the denominator is the derivative of \(g\) at \(a\).

Since \(f\) and \(g\) are assumed to be differentiable at \(a\), the limits in the numerator and the denominator exist and, since \(g'(a)\neq 0\), the limit in the denominator is non-zero.

Therefore, we have \(\displaystyle \lim_{x \rightarrow a} \dfrac{f(x)}{g(x)} = \dfrac{f'(a)}{g'(a)}\).

Since \(f'\) and \(g'\) are continuous at \(a\), we have

\[\begin{align*} \lim_{x \rightarrow a} \dfrac{f'(x)}{g'(x)} & = \frac{\displaystyle \lim_{x\rightarrow a} f'(x)}{\displaystyle \lim_{x\rightarrow a} g'(x)} & \text{limit of the quotient} \\ & = \dfrac{f'(a)}{\underbrace{g'(a)}_{\neq 0}} & f' \text{ and } g' \text{ continuous at } a \\ & = \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \end{align*}\]

as desired.

Lesson Part 4

Examples

Example 2—Part A

Evaluate the limit: \(\displaystyle \lim_{x \rightarrow 1} \frac{\ln(x)}{x-1} \)

Solution

Since \(\ln(1) = 0\) and \(1-1 = 0\), this limit is an indeterminate form of type \(\frac{0}{0}\) (as we saw at the start of this module).

This suggests we should use l'Hospital's Rule to find the limit.

We have \(f(x) = \ln(x)\) and \(g(x) = x-1\), so \(f'(x) = \frac{1}{x}\) and \(g'(x) = 1\).

So here, just like in our special case, \(f'\) and \(g'\) exist and \(g'(x) \neq 0\) for all \(x\). Thus, we can apply l'Hospital's rule:

\[\begin{align*} \lim_{x \rightarrow 1} \frac{\ln(x)}{x-1} & = \lim_{x\rightarrow 1} \frac{~~\frac{1}{x}~~}{~~1~~} \\ & = \lim_{x\rightarrow 1} \frac{1}{x} \\ & = 1 \end{align*}\]

Notice that it does not matter whether we approach \(x=1\) from the left or from the right—we get the same limit in either case using l'Hospital's rule.

This means the function \(\dfrac{\ln(x)}{x-1}\) has a removable discontinuity at the point \(x=1\).

Example 2—Part B

Evaluate the limit: \(\displaystyle \lim_{x \rightarrow \infty} \frac{\ln(x)}{x-1} \)

Solution

As \(x\) approaches infinity, \(\ln(x)\) approaches infinity and \(x-1\) approaches infinity. So in this case, we have an indeterminate form of type \(\frac{\infty}{\infty}\).

Using l'Hospital's rule, we get

\[\begin{align*} \lim_{x \rightarrow \infty} \frac{\ln(x)}{x-1} & = \lim_{x\rightarrow \infty} \frac{~~\frac{1}{x}~~}{~~1~~} \\ & = \lim_{x\rightarrow \infty} \frac{1}{x} \\ & = 0 \end{align*}\]

In the next module, we will use each of these limits to aid in sketching the curve \(y = \dfrac{\ln(x)}{x-1}\).

Example 3—Part A

Determine the trigonometric limit: \(\displaystyle \lim_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\cos(x)}{1-\sin(x)} \)

Solution

Substituting the value \(\dfrac{\pi}{2}\) into the numerator and denominator, we see that \(\cos\left(\dfrac{\pi}{2}\right) = 0\) and \(1-\sin\left(\dfrac{\pi}{2}\right) = 1-1=0\). Therefore, this limit is an indeterminate form of type \(\frac{0}{0}\).

By l'Hospital's rule, we have

\[\begin{align*} \lim_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\cos(x)}{1-\sin(x)} & = \lim_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\dfrac{d}{dx}\Big(\cos(x)\Big)}{\dfrac{d}{dx}\Big(1-\sin(x)\Big)} \\ & = \lim_{x \rightarrow \frac{\pi}{2}^{-}} \frac{-\sin(x)}{-\cos(x)} \\ & = \lim_{x \rightarrow \frac{\pi}{2}^{-}} \tan(x) \rightarrow \infty \end{align*}\]

If we recall the graph of \(\tan(x)\) near \(\dfrac{\pi}{2}\), we'll see that \(\tan(x)\) approaches infinity as \(x\to \frac{\pi}{2}^{-}\).

Example 3—Part B

Determine the trigonometric limit: \(\displaystyle \lim_{x \rightarrow 0} \frac{\sin(x)}{\tan(3x)}\)

Solution

Again, substituting the value of \(x=0\) into the numerator and the denominator, we see that \(\sin(0) = 0\) and \(\tan(3\cdot0) = 0\). Therefore, we have an indeterminate form of type \(\frac{0}{0}\).

By l'Hospital's rule, we have

\[\begin{align*} \lim_{x \rightarrow 0} \frac{\sin(x)}{\tan(3x)} & = \lim_{x \rightarrow 0} \frac{\dfrac{d}{dx}\Big(\sin(x)\Big)}{\dfrac{d}{dx}\Big(\tan(3x)\Big)} \\ & = \lim_{x \rightarrow 0} \frac{\cos(x)}{3\sec^{2}(3x)} &\text{Recall }\frac{d}{du}\tan{u}=\sec^2{u}\text{ and use chain rule} \\ & = \dfrac{\cos(0)}{3 \sec^{2}(3\cdot0)} \\ & = \dfrac{1}{3(1)^{2}} \\ & = \dfrac{1}{3} \\ \end{align*}\]

Lesson Part 5

Examples

Example 4

Let's try an example where l'Hospital's rule is applied to an exponential function.

Determine \(\displaystyle \lim_{x \rightarrow \infty} (1-x^{2})e^{-x}\).

Solution

First, we observe that \(\displaystyle \lim_{x \rightarrow \infty} (1-x^{2} )\rightarrow -\infty\). This is because \(1-x^{2}\) is a parabola that opens downward.

If we recall the graph of \(e^{-x}\), we see that \(\displaystyle \lim_{x \rightarrow \infty} e^{-x} = 0\)

So we have a product of two terms, one approaching \(-\infty\), and one approaching \(0\), and so whether or not this limit exists is not clear.

To use l'Hospital's rule, we need to rewrite this product as a quotient and hope that it's an indeterminate form of type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), which is in fact the case.

\[\lim_{x \rightarrow \infty} (1-x^{2})e^{-x} = \lim_{x \rightarrow \infty} \frac{1-x^{2}}{e^{x}}\]

As the limit in the numerator is \(-\infty\) and the limit in the denominator is \(\infty\), we have an indeterminate form of type \(\frac{\infty}{\infty}\).

Applying l'Hospital's rule, we get

\[\lim_{x \rightarrow \infty} \frac{1-x^{2}}{e^{x}} = \lim_{x \rightarrow \infty} \frac{-2x}{e^{x}}\]

provided that the second limit exists.

Here, it is not clear whether the limit on the right-hand side exists and, if so, what the value is.

Notice that this second limit is also indeterminate.

As \(x\to\infty\), the function \(-2x\) approaches \(-\infty\), and the function \(e^x\) approaches \(\infty\).

Here we have another limit of type \(\frac{\infty}{\infty}\).

So we can apply l'Hospital's rule again, this time to the quotient \(\dfrac{-2x}{e^{x}}\).

\[\begin{align*} \lim_{x \rightarrow \infty} \frac{-2x}{e^{x}} & = \lim_{x \rightarrow \infty} \frac{-2}{e^{x}} \\ & = -2 \lim_{x \rightarrow \infty} \frac{1}{e^{x}} \\ & = -2 \cdot 0 \\ & = 0 \end{align*}\]

Now since the final limit existed, we conclude the following by l'Hospital's rule.

\[\lim_{x \rightarrow \infty} (1-x^{2})e^{-x} = \lim_{x \rightarrow \infty} \frac{-2x}{e^{x}} = \lim_{x \rightarrow \infty} \frac{-2}{e^{x}} = 0\]

Example 5

Determine \(\displaystyle \lim_{x \rightarrow \infty} x \tan\left(\frac{1}{x}\right)\).

Solution

Here we have a product of functions.

We observe that \(\displaystyle \lim_{x \rightarrow \infty} x = \infty\).

As \(x \to \infty\), the term \(\dfrac{1}{x}\) approaches \(0\), and so \(\displaystyle \lim_{x \rightarrow \infty} \tan\left(\frac{1}{x}\right) = \tan(0) = 0\)

So the value of the limit is not clear.

It may not be as obvious as in the previous example, but this is a case where we can actually rewrite this as a quotient and use l'Hospital's rule.

By rewriting the function in the numerator, \(x\), as a function in the denominator, \(\dfrac{1}{x}\), we have

\[\lim_{x \rightarrow \infty} x \tan\left(\frac{1}{x}\right) = \lim_{x \rightarrow \infty} \frac{ \tan\left(\frac{1}{x}\right)}{\frac{1}{x}}\]

Again, the limit in the numerator is \(0\). And now, the limit in the denominator of \(\dfrac{1}{x}\) is equal to \(0\) as well. So this is an indeterminate form of type \(\frac{0}{0}\).

Using l'Hospital's rule, we get

\[\begin{align*} \lim_{x \rightarrow \infty} \frac{ \tan\left(\frac{1}{x}\right)}{\frac{1}{x}} & = \lim_{x \rightarrow \infty} \frac{ \dfrac{d}{dx}\left(\tan\left(\frac{1}{x}\right)\right)}{\dfrac{d}{dx}\left(\frac{1}{x}\right)} \\ & = \lim_{x \rightarrow \infty} \frac{ \sec^{2} \left(\frac{1}{x}\right)\cdot \left(-\frac{1}{x^{2}}\right)}{-\frac{1}{x^{2}}} & \text{by the chain rule} \\ & = \lim_{x \rightarrow \infty} \sec^{2}\left(\frac{1}{x}\right) & \text{by cancelling terms} \\ & = \sec^{2}(0) & \text{as }x\to\infty, \frac{1}{x}\to 0\\ & = (1)^{2} \\ & = 1 \end{align*}\]

Lesson Part 6

Examples

Example 6

Show that \(\displaystyle \lim_{x \rightarrow \infty} \frac{e^{x}}{x^{n}} \rightarrow \infty\) for all positive integers \(n\).

Solution

First we know that as \(x\to\infty\), the exponential \(e^x\) also approaches infinity. Moreover, no matter what positive integer \(n\) we choose, \(\displaystyle \lim_{x \rightarrow \infty} x^n\) approaches infinity. Of course, as \(n\) becomes larger and larger, we would expect the polynomial \(x^n\) to approach infinity at a faster and faster rate. But this limit is showing that no matter how large we pick our integer \(n\), the exponential function in the numerator is always going to dominate. Even though \(e^x\) and \(x^n\) are both approaching infinity, no matter what polynomial we pick, \(e^x\) is always approaching infinity at a faster rate.

Let's prove this using l'Hopital's rule. We fix any positive integer \(n\).

Let \(f(x) = e^{x}\) and \(g(x) = x^{n}\).

Now let's examine the derivatives. \(f\) is an exponential function and so the derivative of \(f\) is the same function \(e^x\), again.

Since \(f'(x) = e^{x}\), we have \(f''(x)= e^{x}\), \(f'''(x) = e^{x}\) and, in general, the \(k^{th}\) derivative \(f^{(k)}(x) = e^{x}\) for all \(k \geq 1\).

On the other hand, we have

\[\begin{align*} g(x) & = x^{n} \\ g'(x) & = n x^{n-1} & \text{by the power rule} \\ g''(x) & = (n-1)n x^{n-2} & \text{by the power rule again} \\ g^{(3)}(x) & = (n-2)(n-1)n x^{n-3} \\ & \ \ \vdots \end{align*}\]

We repeat this process. At each stage, the power of \(x\) is reduced by \(1\). At some point, we must reach \(x^0\). In particular, using a little counting argument, we can show that the \(n-1\)st derivative of \(g\) is equal to some constant times \(x^1\).

\[\begin{align*} g^{(n-1)}(x) & = C~ x^{1} \qquad \text{for some constant } C \\ g^{(n)}(x) & = C \end{align*}\]

Therefore, if we take the derivative one more time, we get the \(n\)th derivative of \(g\) is just equal to that constant \(C\). Now what does this computation of the derivatives tell us about the limit?

Applying l'Hospital's rule \(n\) times, we get the following chain of equalities, provided that each limit exists or approaches \(\infty\) or \(-\infty\):

\[\begin{align*} \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} & = \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)} \\ & = \lim_{x \rightarrow \infty} \frac{f''(x)}{g''(x)} \\ & \ \ \vdots \\ & = \lim_{x \rightarrow \infty} \frac{f^{(n-1)}(x)}{g^{(n-1)}(x)} \\ & = \lim_{x \rightarrow \infty} \frac{f^{(n)}(x)}{g^{(n)}(x)} \end{align*}\]

Well, we know that all the derivatives of the exponential are the same. So \(f^{(n)}(x) = e^x\). But on the other hand, \(g^{(n)}(x)=C\).

\[ \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \infty} \frac{f^{(n)}(x)}{g^{(n)}(x)} = \lim_{x \rightarrow \infty} \frac{e^{x}}{C} \rightarrow \infty \]

This last limit it is easy to compute. It's the limit as \(x\) approaches infinity of the exponential function times some constant. So this limit approaches infinity.

You should be able to use the idea from this example to prove a similar result with the quotient flipped. More specifically, can you show that \(\displaystyle \lim_{x \rightarrow \infty} \frac{x^{n}}{e^{x}} \rightarrow 0\) for all positive integers \(n\)? Try that as an exercise.

Quiz

See the quiz in the side navigation.