Question

$f\left(x\right)=\frac{1+e^{1/x}}{1-e^{1/x}}$ $(x\ne 0)$, $f\left(0\right)=1$, then $f\left(x\right)$ is

• A. left coninuous at $x=0$
• B. right continuous at $x=0$
• C. continuous at $x=0$
• D. none

Answer 1

The correct option is A left coninuous at $x=0$

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

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

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

$={lim}_{x\to 0}\frac{e^{-1/x}+1}{e^{-1/x}-1}=-1\ne f\left(0\right)\Rightarrow$ Not right continues

$\Rightarrow \left(A\right)$