Question

Solve ${lim}_{x\to 0}\frac{\cos x-\cot x}{x}$

Answer 1

$\underset{x\to 0}{lim}$ $\frac{\cos x-\cot x}{x}$

$=\underset{x\to 0}{lim}$ $\frac{\cos x-\frac{\cos x}{\sin x}}{x}$

$=\underset{x\to 0}{lim}$ $\frac{\sin x\cos x-\cos x}{x\sin x}$

$=\underset{x\to 0}{lim}$ $\frac{\cos x\left(\sin x-1\right)}{x\sin x}$

Now ,

$\underset{x\to 0^{-}}{lim}$ $\frac{\cos x\left(\sin x-1\right)}{x\sin x}$

$=\underset{h\to 0}{lim}$ $\frac{\cos \left(0-h\right)\left(\sin \left(0-h\right)-1\right)}{\left(0-h\right)\sin \left(0-h\right)}$

$=\underset{h\to 0}{lim}$ $\frac{\cosh \left(-\sinh -1\right)}{-h\times -\sinh }$

$=\underset{h\to 0}{lim}$ $\frac{-\cosh \left(\sinh +1\right)}{h\sinh }$

$=-\infty$

$\underset{x\to 0^{+}}{lim}$ $\frac{\cos x\left(\sin x-1\right)}{x\sin x}$

$=\underset{h\to 0}{lim}$ $\frac{\cos \left(0+h\right)\left(\sin \left(0+h\right)-1\right)}{\left(0+h\right)\sin \left(0+h\right)}$

$=\underset{h\to 0}{lim}$ $\frac{\cosh \left(\sinh -1\right)}{h\sinh }$

$=\frac{1\times -1}{0}$

$=-\infty$

So ,$\underset{x\to 0}{lim}$ $\frac{\cos x-\cot x}{x}=-\infty$