DETERMINING LIMITS USING L'HOPITAL'S RULES

The following problems involve the use of l'Hopital's Rule. It is used to circumvent the common indeterminate forms $ \frac{ "0" }{ 0 } $ and $ \frac{"\infty" }{ \infty } $ when computing limits. There are numerous forms of l"Hopital's Rule, whose verifications require advanced techniques in calculus, but which can be found in many calculus books. This link will show you the plausibility of l'Hopital's Rule. Following are two of the forms of l'Hopital's Rule.


THEOREM 1 (l'Hopital's Rule for zero over zero): Suppose that $ \displaystyle{ \lim_{x \rightarrow a} f(x) =0 } $, $ \displaystyle{ \lim_{x \rightarrow a} g(x) =0 } $, and that functions $f$ and $g$ are differentiable on an open interval $I$ containing $a$. Assume also that $ g'(x) \ne 0$ in $I$ if $x \ne a$. Then $$ \displaystyle{ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} } = \displaystyle{ \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} } $$ so long as the limit is finite, $+\infty$, or $-\infty$. Similar results hold for $ x \rightarrow \infty $ and $ x \rightarrow - \infty $.


THEOREM 2 (l'Hopital's Rule for infinity over infinity): Assume that functions $f$ and $g$ are differentiable for all $x$ larger than some fixed number. If $ \displaystyle{ \lim_{x \rightarrow a} f(x) = \infty } $ and $ \displaystyle{ \lim_{x \rightarrow a} g(x) = \infty } $, then $$ \displaystyle{ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} } = \displaystyle{ \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} } $$ so long as the limit is finite, $+\infty$, or $-\infty$. Similar results hold for $ x \rightarrow \infty $ and $ x \rightarrow - \infty $.


In both forms of l'Hopital's Rule it should be noted that you are required to differentiate (separately) the numerator and denominator of the ratio if either of the indeterminate forms $ \frac{ "0" }{ 0 } $ or $ \frac{"\infty" }{ \infty } $ arises in the computation of a limit. Do not confuse l'Hopital's Rule with the Quotient Rule for derivatives. Here is a simple illustration of Theorem 1.

EXAMPLE 1: $$ \displaystyle{ \lim_{x \rightarrow 2} \frac{x-2}{x^2-4} = \frac{" 2-2" }{ (2)^2-4 } = \frac{ "0" }{ 0 } } $$ (Now apply Theorem 1. Differentiate top and bottom separately.) $$ = \displaystyle{ \lim_{x \rightarrow 2} \frac{1-0}{2x-0} } $$ $$ = \displaystyle{ \lim_{x \rightarrow 2} \frac{1}{2x} } $$ $$ = \frac{1}{2(2)} $$ $$ = \frac{1}{4} $$ Here is a simple illustration of Theorem 1.

EXAMPLE 2: $$ \displaystyle{ \lim_{x \rightarrow \infty} \frac{2x+7}{3x^2-5} = \frac{"\infty" }{ \infty } = \frac{ "0" }{ 0 } } $$ (Now apply Theorem 2. Differentiate top and bottom separately.) $$ = \displaystyle{ \lim_{x \rightarrow \infty} \frac{2+0}{6x-0} } $$ $$ = \displaystyle{ \lim_{x \rightarrow \infty} \frac{1}{3x} } $$ $$ = \frac{"1"}{\infty} $$ $$ = 0 $$


Indeterminate forms besides $ \frac{ "0" }{ 0 } $ and $ \frac{"\infty" }{ \infty } $ include $ "0 \cdot \infty $, $ "\infty - \infty" $, $ "1^{\infty}" $, $ "0^{0}" $, and $ "\infty^{0}" $. These forms also arise in the computation of limits and can often be algebraically transformed into the form $ \frac{ "0" }{ 0 } $ or $ \frac{"\infty" }{ \infty } $, so that l'Hopital's Rule can be applied. Following are two examples of such transformations. The second example uses the fact that $ y=e^x $ and $ y=\ln x $ are inverse functions, so that $ z= e^{\ln z} $ for all $ z>0 $ and $ \ln z^m=m \ln z$ for all $ z>0 $ and any $ m$.

EXAMPLE 3: $$ \displaystyle{ \lim_{x \rightarrow 0^{+}} \sqrt{x} \cdot \ln x } = " 0 \cdot -\infty"$$ (Circumvent this indeterminate form by "flipping" $ \sqrt{x} $.) $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} \frac{\ln x}{1/\sqrt{x}} } = \frac{ "- \infty" }{ \infty } $$ (Now use Theorem 2 for l'Hopital's Rule.) $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} \frac{1/x}{-1/2x^{3/2}} } $$ $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} -2 \sqrt{x} } $$ $$ = -2 \sqrt{0} $$ $$ = -2 (0) $$ $$ = 0 $$

EXAMPLE 4: $$ \displaystyle{ \lim_{x \rightarrow 0^{+}} x^x } = " 0^0" $$ (Use the fact that $ z=e^{\ln z} $.) $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} e^{ \ln {x^{x}} } } $$ (Use the fact that $ \ln z^m=m \ln z$ .) $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} e^{ x \ln x } } $$ (This next step uses the fact that $y=e^{x \ln x} $ is a continuous function.) $$ = \displaystyle{ e^{\lim_{x \rightarrow 0^{+}} { x \ln x }} } $$ $$ = \displaystyle{ e^{ "0 \cdot (-\infty)"} } $$ (Circumvent this indeterminate form by "flipping" $ x $.) $$ = \displaystyle{ e^{\lim_{x \rightarrow 0^{+}} { \frac{ \ln x }{1/x} } } } $$ (Now apply Theorem 2 for l'Hopita's Rule.) $$ = \displaystyle{ e^{\lim_{x \rightarrow 0^{+}} { \frac{ 1/x }{-1/x^2} } } } $$ $$ = \displaystyle{ e^{\lim_{x \rightarrow 0^{+}} { (-x) } } } $$ $$ = e^0 $$ $$ = 1 $$

In the list of limit problems which follows, most problems are average and a few are somewhat challenging. In some cases there may be methods other than l'Hopital's Rule that could be used to compute the given limit. Nonetheless, l'Hopital's Rule will be used whenever applicable in this problem set.



..... ..... L'HOPITAL ALERT ..... The following problems require algebraic manipulation BEFORE l'Hopital's Rule can be applied.




Click HERE to return to the original list of various types of calculus problems.

Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :

kouba@math.ucdavis.edu

A heartfelt "Thank you" goes to The MathJax Consortium for making the construction of this webpage fun and easy.

Duane Kouba ... May 2, 2016