Question

The right hand and left hand limit of the function are respectively

• A. 1 and 1
• B. 1 and -1
• C. -1 and -1
• D. -1 and 1

Answer 1

Answer: (B)

$LHL=\underset{x\to 0^{-}}{\lim }\left(\frac{e^{1/x}-1}{e^{1/x}+1}\right)$
$=\frac{e^{-\infty }-1}{e^{-\infty }+1}=\frac{0-1}{0+1}=-1$
Dividing both numerator and denominator by $e^{1/x}$, we get
$\underset{x\to 0}{\lim }\left(\frac{1-e^{-1/x}}{1+e^{-1/x}}\right)$
$RHL=\underset{x\to 0^{+}}{\lim }\left(\frac{1-e^{-1/x}}{1+e^{-1/x}}\right)$
$=\frac{1-e^{-\infty }}{1+e^{-\infty }}=\frac{1-0}{1+0}=1$