Question

${lim}_{x\to 0}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$ equal to

Answer 1

We have,

${lim}_{x\to 0}\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$

$\Rightarrow {lim}_{x\to 0}\left(\frac{e^{1/x}(1-\frac{1}{e^{1/x}})}{e^{1/x}(1+\frac{1}{e^{1/x}})}\right)$

$\Rightarrow {lim}_{x\to 0}\left(\frac{1-\frac{1}{e^{1/x}}}{1+\frac{1}{e^{1/x}}}\right)$

$\Rightarrow \frac{1-\frac{1}{e^{1/0}}}{1+\frac{1}{e^{1/0}}}$

$\Rightarrow \frac{1-\frac{1}{e^{\infty }}}{1+\frac{1}{e^{\infty }}}$

$\Rightarrow \frac{1-0}{1+0}$

$\Rightarrow 1$

Hence, this is the answer.