EVALUATING LIMTS USING L'HOPITAL'S RULE


The following problems involve the use of l'Hopital's Rule. It is used to circumvent the common indeterminate forms $\displaystyle{ \lq\lq \ 0 \ '' \over 0 } $ and $\displaystyle{ \lq\lq \ \infty \ '' \over \infty } $ when computing limits. There are numerous forms of l'Hopital's Rule, whose verifications require advanced techniques in calculus, but can be found in most calculus textbooks. Here are two of the forms.

THEOREM 1 (l'Hopital's Rule for zero over zero): Suppose that $ \displaystyle{ \lim_{x \to a} f(x)=0 } ,$ $ \displaystyle{ \lim_{x \to 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 \to a} { f(x) \over g(x) } } $ = $ \displaystyle{ \lim_{x \to a} {f'(x)\over g'(x)} } $ ,

so long as the limit is finite, $ + \infty $, or $ - \infty $. Similar results hold for limits $ x \to \infty $ and $ x \to -\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 \to \infty} f(x)= \infty } $ and $ \displaystyle{ \lim_{x \to \infty} g(x)= \infty } $ , then

$ \displaystyle{ \lim_{x \to \infty} {f(x)\over g(x)}= \displaystyle{ \lim_{x \to \infty} {f'(x)\over g'(x)} } } $ ,

so long as the limit is finite, $ + \infty $, or $ - \infty $. Similar results hold for limits $ x \to -\infty $ and $ x \to a $.

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 $\displaystyle{ \lq\lq \ 0 \ '' \over 0 } $ or $\displaystyle{ \lq\lq \ \infty \ '' \over \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 \to 2} \ {x-2\over x^2-4} } $ = $ \displaystyle{ { \lq\lq \ 2-2 \ '' \over 4-4} } $ = $\displaystyle{ \lq\lq \ 0 \ '' \over 0 } $



(Apply Theorem 1. Differentiate top and bottom separately.)

= $ \displaystyle{ \lim_{x \to 2} \ {1-0 \over 2x-0} } $

= $ \displaystyle{ \lim_{x \to 2} \ {1 \over 2x} } $

= $ \displaystyle{ 1 \over 4 } $ .



Here is a simple illustration of Theorem 2.

EXAMPLE 2 :

$ \displaystyle{ \lim_{x \to \infty} \ {2x+7\over 3x^2-5} } $ = $\displaystyle{ \lq\lq \ \infty \ '' \over \infty } $



(Apply Theorem 2. Differentiate top and bottom separately.) truein

= $ \displaystyle{ \lim_{x \to \infty} \ {2-0 \over 6x-0} } $ = $ \displaystyle{ \lim_{x \to \infty} \ {1 \over 3x } } $

= $ \displaystyle{ \lq\lq \ 1 \ '' \over \infty } $ = $ 0 $ .



Indeterminate forms besides $\displaystyle{ \lq\lq \ 0 \ '' \over 0 } $ and $\displaystyle{ \lq\lq \ \infty \ '' \over \infty } $ include $ \lq\lq \ 0 \cdot \infty \ '' $, $ \lq\lq \ \infty - \infty \ '' $, $ \lq\lq \ 1^{\infty} \ '' $, $ \lq\lq \ 0^{ \ 0} \ '' $, and $ \lq\lq \ \infty ^{ \ 0} \ '' $ . These forms also arise in the computation of limits and can often be algebraically transformed into the form $\displaystyle{ \lq\lq \ 0 \ '' \over 0 } $ or $\displaystyle{ \lq\lq \ \infty \ '' \over \infty } $ so that l'Hopital's Rule can be applied. Following are two examples of such transformations. The second example uses the facts 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 \to 0^+ } \ \sqrt{x} \cdot \ln x } $ = $ \displaystyle{ \lq\lq \ 0 \cdot (- \infty) \ '' } $



(Circumvent this indeterminate form by ``flipping" $ \sqrt{x}$.)

= $ \displaystyle{ \lim_{x \to 0^+ } \ { \ln x \over {1/\sqrt{x}} } } $ = $ \displaystyle{ { \lq\lq \ -\infty \ '' \over \infty } } $



(Now use Theorem 2 for l'Hopital's Rule.)

= $ \displaystyle{ \lim_{x \to 0^+ } \ { {1/x} \over {-1/2x^{3/2}} } } $ = $ \displaystyle{ \lim_{x \to 0^+ } \ { -2 \sqrt{x} } } $ = $ 0 $ .



EXAMPLE 4 :

$ \displaystyle{ \lim_{x \to 0^+ } \ x^x } $ = $ \displaystyle{ \lq\lq \ 0^{ \ 0} \ '' } $



(Use the fact that $ z = e^{\ln z } $ .)

= $ \displaystyle{ \lim_{x \to 0^+ } \ e^{ \ln x^x } } $



(Use the fact that $ \ln z^m = m \ln z $ .)

= $ \displaystyle{ \lim_{x \to 0^+ } \ e^{ x \ln x } } $

= $ \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ {x \ln x } } } } $

= $ \displaystyle{ \ e^ { \ \displaystyle{ \lq\lq \ 0 \cdot (-\infty) \ '' } } } $



(Circumvent this indeterminate form by ``flipping" $ x $.)

= $ \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { \ln x \over {1/x} } } } } $ = $ \displaystyle{ \ e^ { \ \displaystyle{ { \lq\lq \ \ln 0 / \infty \ '' } } } } $ = $ \displaystyle{ \ e^ { \ \displaystyle{ { \lq\lq -\infty / \infty \ '' } } } } $



(Now use Theorem 2 for l'Hopital's Rule.)

= $ \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { {1/x} \over {-1/x^2} } } } } $

= $ \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ (-x) } } } $

= $ \displaystyle{ \ e^ { \ 0} } $

= $ 1 $ .

In the list of 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 a given limit. Nonetheless, l'Hopital's Rule will be used wherever applicable in this problem set.