Question

The value of $\underset{x\to 0}{lim}\frac{e^{1/x}-1}{e^{1/x}+1}$ is equal to

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

Answer 1

The correct option is A $1$

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

$\underset{x\to 0}{lim}\frac{e^{1/x}+1-2}{e^{1/x}+1}$

$\underset{x\to 0}{lim}\frac{e^{1/x}+1}{e^{1/x}+1}=\frac{2}{e^{1/x}+1}$

$=\underset{x\to 0}{lim}1-\underset{x\to 0}{lim}\frac{2}{e^{1/x}+1}$

$=1-\underset{x\to 0}{lim}\frac{2}{e^{1/x}+1}=1-\frac{2}{e^{\infty }+1}\to$ tends to O (as denominator tends to $\infty$)

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