Indeterminate Forms Involving Transcendental Functions

Find the limits.

  1. \displaystyle \lim_{x\rightarrow1^{+}}\left(\frac{1}{\ln x}-\frac{1}{\cosh^{-1}x}\right)
  2. \displaystyle \lim_{x\rightarrow0^{+}}\left(e^{x}+\sinh x\right)^{{\displaystyle \cot x}}

Solution

Note that we shall use l’Hospital’s Rule in our computations as necessary.

The first limit has the indeterminate form \infty-\infty. We compute the limit as follows:

\displaystyle \lim_{x\rightarrow 1^{+}}\left(\frac{1}{\ln x}-\frac{1}{\cosh^{-1}x}\right) = \lim_{x\rightarrow 1^{+}}\left(\frac{\cosh^{-1}x-\ln x}{\ln x\cdot\cosh^{-1}x}\right)\: \mathrm{Form}:\frac{0}{0}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow 1^{+}} \left( \frac{1}{\ln x} - \frac{1}{\cosh^{-1}x}\right)} = \lim_{x\rightarrow 1^{+}}\left(\frac{\dfrac{1}{\sqrt{x^{2}-1}} - \dfrac{1}{x}}{\dfrac{\ln x}{\sqrt{x^{2}-1}} + \dfrac{\cosh^{-1}x}{x}}\right)

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow 1^{+}} \left( \frac{1}{\ln x} - \frac{1}{\cosh^{-1}x}\right)} = \lim_{x\rightarrow 1^{+}}\frac{x-\sqrt{x^{2}-1}}{x\ln x+\cosh^{-1}x\cdot\sqrt{x^{2}-1}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow 1^{+}} \left( \frac{1}{\ln x} - \frac{1}{\cosh^{-1}x}\right)} = \frac{1}{0^{+}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow 1^{+}} \left( \frac{1}{\ln x} - \frac{1}{\cosh^{-1}x}\right)} = +\infty

Meanwhile, the second limit has the form 1^{\infty}. Observe that

\displaystyle \lim_{x\rightarrow 0^{+}}\cot x\cdot\ln\left(e^{x}+\sinh x\right) = \lim_{x\rightarrow 0^{+}}\frac{\ln\left(e^{x}+\sinh x\right)}{\tan x}\:\left(\mathrm{Form}:\frac{0}{0}\right)

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x_\rightarrow 0^{+}}\cot x\cdot \ln \left( e^{x} +\sinh x\right)} = \lim_{x\rightarrow 0^{+}} \frac{\left(\dfrac{e^{x}+\cosh x}{e^{x}+\sinh x}\right)}{\sec^{2}x}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x_\rightarrow 0^{+}}\cot x\cdot \ln \left( e^{x} +\sinh x\right)} = 2.

Therefore,

\displaystyle \lim_{x\rightarrow0^{+}}\left(e^{x}+\sinh x\right)^{{\displaystyle \cot x}} = e^{\displaystyle \left[ \lim_{x\rightarrow 0^{+}}\ln\left(e^{x}+\sinh x\right)^{\cot x}\right]}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow0^{+}}\left(e^{x}+\sinh x\right)^{{\displaystyle \cot x}}} = e^{\displaystyle \left[ \lim_{x\rightarrow 0^{+}}\cot x\cdot\ln\left(e^{x}+\sinh x\right)\right]}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow0^{+}}\left(e^{x}+\sinh x\right)^{{\displaystyle \cot x}}} = e^{2}

This entry was posted on Sunday, March 23rd, 2008 at 1:32 am and is filed under Calculus. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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