Applications of L'Hopital's Rule

From Indeterminate Forms to Useful Limits

The previous tutorial considered how to recognize and transform indeterminate forms. In this tutorial, the focus is on what L’Hopital’s Rule can accomplish once its hypotheses are satisfied: it can resolve repeated cancellation, simplify quotients whose terms have different growth rates, and establish general comparisons among logarithms, powers, and exponentials.

The rule is not a license to differentiate any difficult quotient. At each application, the quotient must still have the required indeterminate form, the functions must satisfy the relevant differentiability conditions, and the derivative of the denominator must be nonzero near the endpoint. Repeated applications require checking these conditions again for the new quotient.

Repeated Applications at a Finite Endpoint

Sometimes both a numerator and denominator vanish to several orders at an endpoint. If their derivatives continue to vanish there, L’Hopital’s Rule may be applied successively. The following corollary records one useful version of that procedure. It follows from the zero-over-zero form of L’Hopital’s Rule established earlier in this course.

Theorem (Repeated Zero-Over-Zero Rule): Let $n$ be a positive integer, and suppose $f$ and $g$ have derivatives through order $n$ on $(a,a+\delta)$. For each $k=0,\ldots,n-1$, suppose $f^{(k)}$ and $g^{(k)}$ extend continuously to $a$, with $f^{(k)}(a)=g^{(k)}(a)=0$. Suppose also that $g^{(k)}(x)\ne0$ for $k=0,\ldots,n$ and $a<x<a+\delta$. If $$ \lim_{x\to a^+}\frac{f^{(n)}(x)}{g^{(n)}(x)}=L, $$ where $L$ is finite or infinite, then $$ \lim_{x\to a^+}\frac{f(x)}{g(x)}=L. $$

Proof. For each $k=0,\ldots,n-1$, the functions $f^{(k)}$ and $g^{(k)}$ are continuous at $a$, differentiable on $(a,a+\delta)$, and both have value zero at $a$. Their quotient is defined for $x>a$ because $g^{(k)}(x)\ne0$. The derivative of the denominator is $g^{(k+1)}(x)$, which is nonzero there by hypothesis. Thus the zero-over-zero form of L’Hopital’s Rule gives

\lim_{x\to a^+}\frac{f^{(k)}(x)}{g^{(k)}(x)} = \lim_{x\to a^+}\frac{f^{(k+1)}(x)}{g^{(k+1)}(x)}

whenever the limit on the right exists. Starting with $k=n-1$, the assumed limit of $f^{(n)}/g^{(n)}$ therefore gives the limit of $f^{(n-1)}/g^{(n-1)}$. Applying the same implication for $k=n-2$, then $k=n-3$, and continuing down to $k=0$, gives the stated limit of $f/g$. The argument applies whether $L$ is finite or infinite, as allowed by the rule. $\square$

Worked Example: A Quotient with Two Orders of Cancellation

Evaluate

\lim_{x\to0}\frac{e^{2x}-1-2x}{x^2}.

Both numerator and denominator tend to zero. Their first derivatives are $2e^{2x}-2$ and $2x$, respectively, and both also tend to zero. Their second derivatives are $4e^{2x}$ and $2$. The denominator and its first two derivatives are nonzero for $x\ne0$ sufficiently near zero: the denominator is $x^2$, its first derivative is $2x$, and its second derivative is $2$. Applying L’Hopital’s Rule twice gives

\lim_{x\to0}\frac{e^{2x}-1-2x}{x^2} = \lim_{x\to0}\frac{2e^{2x}-2}{2x} = \lim_{x\to0}\frac{4e^{2x}}{2} =2.

For a two-sided limit, the same calculation is valid on both sides of zero: the required zero-over-zero conditions hold as $x\to0^+$ and as $x\to0^-$. Both one-sided limits are $2$, so the two-sided limit is $2$ as well. The repeated rule explains why two applications are justified here: the first derivative quotient is still of the zero-over-zero form.

Repeated Applications at Infinity

Repeated differentiation is also useful when both functions tend to infinity. Each step must produce another quotient to which the infinity-over-infinity form of L’Hopital’s Rule applies. The following example keeps track of the form at every stage.

Worked Example: Comparing Polynomials of the Same Degree

Find

\lim_{x\to\infty}\frac{5x^3-2x+1}{2x^3+7x^2}.

The original numerator and denominator both tend to $+\infty$. The derivative of the denominator is $6x^2+14x$, which is nonzero for $x>0$. One application gives

\lim_{x\to\infty} \frac{15x^2-2}{6x^2+14x}.

This is still an infinity-over-infinity form. The next denominator derivative, $12x+14$, is nonzero for $x>0$, and another application gives

\lim_{x\to\infty}\frac{30x}{12x+14}.

Again, both numerator and denominator tend to $+\infty$, and the next denominator derivative is $12$, which is nonzero. A third application yields

\lim_{x\to\infty}\frac{30}{12}=\frac{5}{2}.

Thus the original limit is $5/2$. At each step the quotient remains in the required form, and the denominator derivative is nonzero. The calculation illustrates a general technique for polynomial quotients: if several derivative quotients are still indeterminate, keep applying the rule only while its conditions remain satisfied.

Comparing Logarithms and Powers

L’Hopital’s Rule can establish growth comparisons that apply to many later limits. A logarithm grows without bound as its input tends to infinity, but it grows more slowly than every positive power of that input.

Theorem (Logarithmic Growth Is Slower Than Every Positive Power): For every real $p>0$, $$ \lim_{x\to\infty}\frac{\ln x}{x^p}=0. $$

Proof. As $x\to\infty$, both $\ln x$ and $x^p$ tend to $+\infty$, so the quotient has the infinity-over-infinity form. For $x>0$, the derivative of the denominator is $p x^{p-1}$, which is nonzero because $p>0$. L’Hopital’s Rule gives

\lim_{x\to\infty}\frac{\ln x}{x^p} = \lim_{x\to\infty} \frac{1/x}{p x^{p-1}} = \lim_{x\to\infty}\frac{1}{p x^p} =0.

The last equality follows because $p x^p\to+\infty$. This proves the claimed limit. $\square$

Worked Example: A Squared Logarithm Compared with a Power

Evaluate

\lim_{x\to\infty}\frac{(\ln x)^2}{x}.

The numerator and denominator both tend to $+\infty$. The derivative of the denominator is $1$, which is nonzero, so L’Hopital’s Rule gives

\lim_{x\to\infty}\frac{(\ln x)^2}{x} = \lim_{x\to\infty}\frac{2\ln x}{x}.

The new numerator and denominator also tend to $+\infty$, so the rule applies again. Differentiating gives

\lim_{x\to\infty}\frac{2\ln x}{x} = \lim_{x\to\infty}\frac{2/x}{1} = \lim_{x\to\infty}\frac{2}{x} =0.

Therefore the original limit is zero. Notice the role of the second application: after the first differentiation, the expression was still an indeterminate quotient. The theorem on logarithmic growth also gives the same conclusion by writing the quotient as $\ln x/ \sqrt{x}$ times itself only with further care; the direct repeated calculation makes the required steps explicit.

Exponentials and Polynomial Growth

The same method establishes an even stronger comparison: the exponential function eventually grows faster than every fixed-degree polynomial. Here the degree is a nonnegative integer, so repeated differentiation eventually reduces the polynomial numerator to a constant.

Theorem (Exponential Growth Dominates Polynomial Growth): For every nonnegative integer $n$, $$ \lim_{x\to\infty}\frac{x^n}{e^x}=0. $$

Proof. If $n=0$, the quotient is $1/e^x$, which tends to zero as $x\to\infty$. Now suppose $n\geq1$. For each $k=0,\ldots,n-1$, the $k$th derivative of $x^n$ is a positive constant times $x^{n-k}$, so it tends to $+\infty$; the $k$th derivative of $e^x$ is $e^x$, which also tends to $+\infty$. The derivative of the denominator is always $e^x\ne0$. Therefore L’Hopital’s Rule applies at each of these $n$ stages. After $n$ differentiations, the numerator is the constant $n!$, and the denominator remains $e^x$. Consequently,

\lim_{x\to\infty}\frac{x^n}{e^x} = \lim_{x\to\infty}\frac{n!}{e^x} =0.

This proves the result for every $n$. $\square$

Worked Example: A High-Degree Polynomial over an Exponential

For a specific application, consider

\lim_{x\to\infty}\frac{3x^4+2x}{e^x}.

Both numerator and denominator tend to $+\infty$, and the denominator derivative $e^x$ is always positive. After one application the quotient is $(12x^3+2)/e^x$, again an infinity-over-infinity form. After a second it is $(36x^2)/e^x$; after a third it is $72x/e^x$; after a fourth it is $72/e^x$. Each intermediate numerator tends to $+\infty$, and the denominator remains $e^x$, so all four applications are valid. The final limit is zero:

\lim_{x\to\infty}\frac{3x^4+2x}{e^x} = \lim_{x\to\infty}\frac{72}{e^x} =0.

The general theorem gives this conclusion term by term as well: $3x^4/e^x\to0$ and $2x/e^x\to0$, so their sum tends to zero.

When the Rule Helps—and When It Does Not

These applications show two particularly effective uses of L’Hopital’s Rule: successive differentiation when cancellation persists, and quantitative comparisons of growth rates. In both settings, the rule is a way to prove a limit, not a substitute for checking what form the quotient has.

Check the original form. The quotient must be of the $0/0$ or $\infty/\infty$ type before an application.
Check the new form. After differentiating, confirm that the derivative quotient has a limit or is still a permitted indeterminate form before applying the rule again.
Check the denominator derivative. It must be nonzero on the relevant punctured interval or tail, as required by the version of the rule being used.
Do not confuse a derivative quotient with the original quotient. L’Hopital’s Rule relates their limits under its hypotheses; it does not assert that the quotients are equal at each point.

A common error is to differentiate repeatedly just because the first derivative quotient remains complicated. Complexity is not the criterion: the next quotient must still satisfy the hypotheses. Conversely, when the quotient remains $0/0$ or $\infty/\infty$ and the derivative conditions hold, another application can be entirely justified.

Check Your Understanding

Use the hypotheses and applications in this tutorial to answer the following questions.

In the repeated zero-over-zero rule, what endpoint conditions must hold for each successive pair of derivatives?
Why is a second application valid in the limit of $(\ln x)^2/x$?
For $p>0$, what quotient results from differentiating the numerator and denominator of $\ln x/x^p$?
How many applications of L’Hopital’s Rule reduce $x^n/e^x$ to a constant over $e^x$ when $n\geq1$?
What must be checked before applying L’Hopital’s Rule to a derivative quotient for a second time?