Question

$\underset{x\to 0}{lim}xlo{g}_e\left(sinx\right)$ is equal to

• A. -1
• B. $lo{g}_{e}1$
• C. 1
• D. none of these

Answer 1

Answer: (B)

$lo{g}_{e}1$

We have,
$\underset{x\to 0}{lim}xlo{g}_{e}sinx$ $[0\times \infty form]$
$=\underset{x\to 0}{lim}\frac{lo{g}_{e}sinx}{\frac{1}{x}}$ $\left[\frac{\infty }{\infty }form\right]$

$=\underset{x\to 0}{lim}\frac{cotx}{\frac{-1}{x^2}}$ [Using L' Hospital's rule]

$=-\underset{x\to 0}{lim}\frac{x^2}{tanx}=-\underset{x\to 0}{lim}\frac{x}{tanx}\times x=-(1\times 0)=0=lo{g}_{e}1$