Evaluate the limit \lim_{x\rightarrow 0^+}(1/tan(x))^{1/ln(x)}?

Feb 11, 2018

$\frac{1}{e}$

Explanation:

let $y = {\lim}_{x \rightarrow {0}^{+}} \left({\left(\frac{1}{\tan} \left(x\right)\right)}^{\frac{1}{\ln} \left(x\right)}\right)$

take ln() of both sides: $\ln y = \ln \left[{\lim}_{x \rightarrow {0}^{+}} \left({\left(\frac{1}{\tan} \left(x\right)\right)}^{\frac{1}{\ln} \left(x\right)}\right)\right]$

move ln() into the limit: $\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(\ln \left[{\left(\frac{1}{\tan} \left(x\right)\right)}^{\frac{1}{\ln} \left(x\right)}\right]\right)$

logarithm properties: $\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(\left(\frac{1}{\ln} \left(x\right)\right) \ln \left(\frac{1}{\tan} \left(x\right)\right)\right)$
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(\ln \frac{\frac{1}{\tan} \left(x\right)}{\ln} \left(x\right)\right)$

if you plug in $x = {0}^{+}$ directly, you get: $\ln \frac{\frac{1}{\tan} \left({0}^{+}\right)}{\ln} \left({0}^{+}\right)$
$= \ln \frac{\frac{1}{0} ^ +}{\ln} \left({0}^{+}\right)$
$= \ln \frac{\infty}{-} \infty$
$= \frac{\infty}{-} \infty$
indeterminate, so use l'hopital's rule:

$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(\frac{\frac{d}{\mathrm{dx}} \left[\ln \left(\frac{1}{\tan} \left(x\right)\right)\right]}{\frac{d}{\mathrm{dx}} \left[\ln \left(x\right)\right]}\right)$
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(\frac{\tan \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{\tan} \left(x\right)\right)}{\frac{1}{x}}\right)$
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(x \tan \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(\cot \left(x\right)\right)\right)$
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(x \tan \left(x\right) \cdot \left(- {\csc}^{2} \left(x\right)\right)\right)$
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(\frac{- x \sin \frac{x}{\cos} \left(x\right)}{\sin} ^ 2 \left(x\right)\right)$
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(- \frac{x}{\cos \left(x\right) \sin \left(x\right)}\right)$

again, plugging in $x = {0}^{+}$ results in 0/0 (indeterminate), so use l'hopitals rule again:
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(\frac{\frac{d}{\mathrm{dx}} \left[- x\right]}{\frac{d}{\mathrm{dx}} \left[\left(\cos \left(x\right) \sin \left(x\right)\right)\right]}\right)$
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(- \frac{1}{- {\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)}\right)$

now plug in $x = {0}^{+}$:
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(- \frac{1}{- {\sin}^{2} \left({0}^{+}\right) + {\cos}^{2} \left({0}^{+}\right)}\right)$
$\ln y = {\lim}_{x \rightarrow {0}^{+}} \left(- \frac{1}{1}\right)$
$\ln y = - 1$
$y = {e}^{- 1}$
$y = \frac{1}{e}$

you've solved for y, which equals the limit you want to find.

check with this graph: graph{(1/tan(x))^(1/ln(x)) [-1.695, 2.15, -0.3, 1.742]}

as x approaches 0 from the right, the limit is about 0.37, or $\frac{1}{e}$