Question

Evaluate the limit :-

$\displaystyle \lim_{x\:to\:0^-} \frac{ {e}^{ \frac{1}{x} } + 1 }{ {e}^{ \frac{1}{x} } - 1}$​

Answer 1

GiveN

The function $\frac{ {e}^{ \frac{1}{x} } + 1 }{ {e}^{ \frac{1}{x} } - 1}$ .

To Evaluate

The limit when x tends to 0.

$\displaystyle \lim_{x\:to\:0^-} \frac{ {e}^{ \frac{1}{x} } + 1 }{ {e}^{ \frac{1}{x} } - 1}$

SolutioN

$\displaystyle \lim_{x \: to \: {0}^{ - } } \frac{1}{x}$ = - $\infty$

$\displaystyle \lim_{x \: to \: {0}^{ - } } {e}^{ \frac{1}{x} } = 0 \: \: \: (since \: {e}^{ - \infty } = 0)$

$\displaystyle \lim_{x\:to\:0^-} \frac{ {e}^{ \frac{1}{x} } + 1 }{ {e}^{ \frac{1}{x} } - 1}$ = $\displaystyle \lim_{x\:to\:0^-} \frac{ 0 + 1 }{ 0 - 1}$ = - 1 Ans.

$\boxed{ \green { Hence, \displaystyle \lim_{x\:to\:0^-} \frac{ {e}^{ \frac{1}{x} } + 1 }{ {e}^{ \frac{1}{x} } - 1} = -1 } }$