A) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\]does not exist
B) \[f(x)\]is continuous at \[x=0\]
C) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=1\]
D) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\]exists but \[f(x)\]is not continuous at \[x=0\]
Correct Answer: D
Solution :
Given \[f(x)=\left\{ \begin{align} & \frac{{{e}^{\frac{1}{x}}}-1}{{{e}^{\frac{1}{x}}}+1}\,\,,\,\,x\ne 0 \\ & 0\,\,\,\,\,\,\,\,\,\,\,\,\,,\,x=0 \\ \end{align} \right.\] Þ \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{e}^{\frac{1}{x}}}-1}{{{e}^{\frac{1}{x}}}+1}=\frac{{{e}^{\infty }}-1}{{{e}^{\infty }}+1}=-1\] Þ \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{e}^{\frac{1}{x}}}-1}{{{e}^{\frac{1}{x}}}+1}=\frac{1-{{e}^{-\frac{1}{x}}}}{1+{{e}^{\frac{1}{x}}}}=\frac{1-{{e}^{-\infty }}}{1+{{e}^{\infty }}}=1\] So, \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\] exists at \[x=0\], but at \[x=0\] it is not continuous.
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