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Show that the function $f(x)=\begin{cases}\frac{e^{1/x}-1}{e^{1/x}+1}, \:x\neq0\\0, \quad\:\:\: x=0\end{cases}$ is discontinuous at $x=0$.

My Attempt

From the graph of the function it is clearly discontinuous at $x=0$, enter image description here

$$ f(0)=0\\ LHL=\lim_{x\to 0^-}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}=\\ RHL=\lim_{x\to 0^+}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}=\\ $$ How do I proceed further and prove $LHL\neq RHL$ ?

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  • $\begingroup$ evaluate the limits: LHL = -1, RHL = 1 QED LHL≠RHL $\endgroup$ – user29418 Apr 2 '18 at 4:30
  • $\begingroup$ There’s no reason to bother showing that LHL$\ne$RHL, as long as you know that one of these is of absolute value $1$, since it’s been given to you that $f(0)=0$. For small positive numbers, the formula is clearly of form $\frac{H-1}{H+1}$, where $H$ stands for a huge positive number. Try $x=1/10$, for instance. $\endgroup$ – Lubin Apr 2 '18 at 4:43
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With $x_{n}=1/n$, then $f(x_{n})=\dfrac{e^{n}-1}{e^{n}+1}=\dfrac{1-e^{-n}}{1+e^{-n}}\rightarrow\dfrac{1-0}{1+0}=1$.

With $y_{n}=-1/n$, then $f(y_{n})=\dfrac{e^{-n}-1}{e^{-n}+1}\rightarrow\dfrac{0-1}{0+1}=-1$.

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$\dfrac{e^{1/x}-1}{e^{1/x}+1} =\dfrac{e^{1/x}+1-1-1}{e^{1/x}+1} =1-\dfrac{2}{e^{1/x}+1} $ and $\lim_{x \to 0+} \dfrac{2}{e^{1/x}+1} =0 $ and $\lim_{x \to 0-} \dfrac{2}{e^{1/x}+1} =2 $.

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