Question

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

​(a) $-\frac{1}{2}$

(b) 1

(c) $\frac{1}{2}$

(d) 1

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to 0\end{array}\frac{\mathrm{cosec}x-\cot x}{x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{{\displaystyle \frac{1}{\sin x}}-{\displaystyle \frac{\cos x}{\sin x}}}{x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{1-\cos x}{x\sin x} \\ \left[\mathrm{since}1-\cos 2\theta =2{\sin }^2\theta \mathrm{and}\sin 2\theta =2\sin \theta \cos \theta \right] \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{2{\sin }^2{\displaystyle \frac{x}{2}}}{x2\cos {\displaystyle \frac{x}{2}}\sin {\displaystyle \frac{x}{2}}} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sin {\displaystyle \frac{x}{2}}}{x}\times \frac{1}{\cos {\displaystyle \frac{x}{2}}} \\ \begin{array}{c}\lim \\ x\to 0\end{array}\left(\frac{\sin {\displaystyle \frac{x}{2}}}{2{\displaystyle \frac{x}{2}}}\right)\left(\begin{array}{c}\lim \\ x\to 0\end{array}\frac{1}{\cos {\displaystyle \frac{x}{2}}}\right) \\ =\frac{1}{2}\left(\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\sin {\displaystyle \frac{x}{2}}}{{\displaystyle \frac{x}{2}}}\right)\left(\frac{1}{1}\right) \\ =\frac{1}{2}\left(\because \begin{array}{c}\lim \\ \theta \to 0\end{array}\frac{\sin \theta }{\theta }=1\right) \end{gathered}$$

Hence,the correct answer is option C.