Question

Evaluate the given limit :
$\underset{x\to 0}{lim}(\text{cosec}x-\cot x)$

Answer 1

Given : $\underset{x\to 0}{lim}(\text{cosec}x-\cot x)$
$\underset{x\to 0}{lim}(\text{cosec}x-\cot x)$
Substituting $x=0$ in the given limit,
$=\text{cosec}0-\cot 0$
$=\infty -\infty$
Since it is in $\infty -\infty$ form.

We need to simplify it,
$\underset{x\to 0}{lim}(\text{cosec}x-\cot x)$
Let $L=\underset{x\to 0}{lim}(\frac{1}{\sin x}-\frac{\cos x}{\sin x})$
$\Rightarrow L=\frac{1-\cos x}{\sin x}$
Substituting $x=0$
$\Rightarrow L=\frac{1-\cos 0}{\sin 0}$
$\Rightarrow L=\frac{1-1}{0}$
$\Rightarrow L=\frac{0}{0}$
It is form $\frac{0}{0}$
Hence,
$\Rightarrow L=\underset{x\to 0}{lim}\frac{1-\cos x}{\sin x}$
Multiplying and dividing by $(1+\cos x)$
$L=\underset{x\to 0}{lim}\frac{1-\cos x}{\sin x}\times \frac{1+\cos x}{1+\cos x}$
$\Rightarrow L=\underset{x\to 0}{lim}\frac{1^2-{\cos }^{2}x}{\sin x(1+\cos x)}$
$\Rightarrow L=\underset{x\to 0}{lim}\frac{1-{\cos }^{2}x}{\sin x(1+\cos x)}$
$\Rightarrow L=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{\sin x(1+\cos x)}$
$\Rightarrow L=\underset{x\to 0}{lim}\frac{\sin x}{1+\cos x}$
Substituting $x=0$
$\Rightarrow L=\frac{\sin 0}{(1+\cos 0)}$
$\Rightarrow L=\frac{0}{1+1}$
$\Rightarrow L=\frac{0}{2}$
$\Rightarrow L=0$