Question

Limits of the form tends to infinite ${lim}_{x\to 0}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}=$

• A. $1$
• B. $-1$
• C. $0$
• D. does not exists

Answer 1

The correct option is A $1$

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

Divide numerator and denominator by $e^{1/x}$

$={lim}_{x\to 0}\frac{\frac{e^{1/x}}{e^{1/x}}-\frac{1}{e^{1/x}}}{\frac{e^{1/x}}{e^{1/x}}+\frac{1}{e^{1/x}}}$

$={lim}_{x\to 0}\frac{1-e^{-1/x}}{1+e^{1/x}}$

as we know ${lim}_{x\to 0}{e}^{-1/x}=e^{-\infty }=0$

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

$=1$

Hence, the answer is $1.$