What is the limit as x approaches 0 of cotx/lnx?

Dec 11, 2014

Since $\ln x$ is only defined when $x > 0$, we can only talk about the right-hand limit:

${\lim}_{x \to {0}^{+}} \frac{\cot x}{\ln x}$

Since

$\left\{\begin{matrix}{\lim}_{x \to {0}^{+}} \cot x = {\lim}_{x \to {0}^{+}} \frac{\cos x}{\sin x} = \frac{1}{{0}^{+}} = + \infty \\ {\lim}_{x \to {0}^{+}} \ln x = - \setminus \infty\end{matrix}\right.$,

by applying l'H$\hat{\text{o}}$pital's Rule ($\infty$/$\infty$),

$= {\lim}_{x \to {0}^{+}} \frac{- {\csc}^{2} x}{\frac{1}{x}}$

by $\csc x = \frac{1}{\sin x}$,

$= - {\lim}_{x \to {0}^{+}} \frac{x}{{\sin}^{2} x}$

by l'H$\hat{\text{o}}$pital's Rule ($0 \text{/} 0$),

$= - {\lim}_{x \to {0}^{+}} \frac{1}{2 \sin x \cos x} = - \frac{1}{0} ^ + = - \infty$

The graph of $y = \frac{\cot x}{\ln x}$ looks like:

As shown above, as $x$ approaches 0 from the right, the graph is shooting down toward $- \infty$.