The function `f(x)={(e^(1/x)-1)/(e^(1/x)+1),x!=0 0,x=0` (a)is continuous at `x=0` (b)is not continuous at `x=0` (c)is not continuous at `x=0,` but can be made continuous at `x=0` (d) none of these

Answer

Option C is correct

Explanation

$$
\begin{aligned}
&f(x)=\frac{e^{1 / x}-1}{e^{1 / x}+1} \\
&f(x)_{x \rightarrow 0^{-}}=\lim _{h \rightarrow 0} \frac{e^{-1 / h}-1}{e^{-1 / h}+1} \\
&f(x)_{x \rightarrow 0^{-}}=\lim _{h \rightarrow 0} \frac{\frac{1}{e^{1 / h}}-1}{\frac{1}{e^{1 / h}}+1}=\frac{0-1}{0+1}=-1 \\
&f(x)_{x \rightarrow 0^{+}}=\lim _{h \rightarrow 0} \frac{e^{1 / h}-1}{e^{1 / h}+1} \\
&f(x)_{x \rightarrow 0^{+}}=\lim _{h \rightarrow 0} \frac{1-\frac{1}{e^{1 / h}}}{1-\frac{1}{e^{1 / h}}}=\frac{1-0}{1+0}=1 \\
&f(x)_{x \rightarrow 0^{-}} \ne f(x)_{x \rightarrow 0^{+}}
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