Question

Evaluate: $\underset{x\to 0}{lim}\frac{x(e^{\sin x}-1)}{1-\cos x}$

• A. $\frac{1}{2}$
• B. $2$
• C. $0$
• D. $1$

Answer 1

The correct option is B $2$

We know $\underset{x\to 0}{lim}=\frac{e^x-1}{x}=1$

Using this

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

$=\underset{x\to 0}{lim}x\sin x\left(\frac{e^{\sin x}-1}{\sin x}\right)\cdot \frac{1}{2{\sin }^2\frac{x}{2}}$(since $1-\cos x=2{\sin }^2\frac{x}{2}$)

$=\underset{x\to 0}{lim}\frac{x\cdot 2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}}\frac{e^{\sin x}-1}{\sin x}$ Since $(\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$)

$=\underset{x\to 0}{lim}\frac{\cos x/2}{\frac{\sin \frac{x}{2}}{x}}=\frac{e^{\sin x}-1}{\sin x}$

$=\underset{x\to 0}{lim}\frac{\cos \left(\frac{x}{2}\right)}{\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}\times 2}}\frac{e^{\sin x}-1}{\sin x}$

$=2\underset{x\to 0}{lim}\frac{(\cos x/2}{\frac{\sin \left(x/2\right)}{x/2}}\frac{e^{\sin x}-1}{\sin x}=2\times \frac{1}{1}\times 1$

$=2$.