MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{2}{x}\log (1 + x) = . . .$
- A$e$
- B${e^2}$
- C$\frac{1}{2}$
- ✓$2$
$ = \mathop {\lim }\limits_{x \to 0} 2{\log _e}e = 2$
$\left\{ { \because \mathop {\lim }\limits_{x \to 0} {{(1 + x)}^{\frac{1}{x}}} = {{\log }_e}e = 1} \right\}$
Trick : Using $L$ Hospital’s rule.
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