Question

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

• A. -1/2
• B. 1/2
• C. 1
• D. 3/2

Answer 1

The correct option is B 1/2

Given,

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

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

Apply L-Hospital's rule

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

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

Apply L-Hospital's rule

$=\frac{1}{2}\cdot \underset{x\to 0}{lim}\left(\frac{\frac{3e^{\sin \left(x\right)}\sin \left(2x\right)+2e^x-2e^{\sin \left(x\right)}{\cos }^3\left(x\right)+2e^{\sin \left(x\right)}\cos \left(x\right)}{2}}{\cos \left(x\right)}\right)$

$=\frac{1}{2}\cdot \underset{x\to 0}{lim}\left(\frac{3e^{\sin \left(x\right)}\sin \left(2x\right)+2e^x-2e^{\sin \left(x\right)}{\cos }^3\left(x\right)+2e^{\sin \left(x\right)}\cos \left(x\right)}{2\cos \left(x\right)}\right)$

$=\frac{1}{2}\cdot \frac{3e^{\sin \left(0\right)}\sin (2\cdot 0)+2e^0-2e^{\sin \left(0\right)}{\cos }^3\left(0\right)+2e^{\sin \left(0\right)}\cos \left(0\right)}{2\cos \left(0\right)}$

$=\frac{1}{2}$