I have to prove that $$\lim_{x\to0}\frac{e^{1/x}-1}{e^{1/x}+1}$$
does not exist, using epsilon-delta.
My attempt: We first consider left hand limit. Let $\varepsilon$ be given.
Now, choosing $\delta=\varepsilon/2$
$$\left|\frac{e^{1/x}-1}{e^{1/x}+1)-1}\right|=2\left|\frac{e^{1/x}}{e^{1/x}+1}\right| <2x <2\delta< \varepsilon$$
Is this correct? And I have no clue how to prove right hand limit. I know it is 1, but how do I prove with epsilon-delta method?
Thanks in advance.
Best Answer
Without loss of generality choose $0< \epsilon <1$ and take $\delta =[\ln(\frac{2}{\epsilon}-1)]^{-1}>0$. If $x \in (-\delta,0)$
\begin{align} &x > - \left[\ln \left(\frac{2}{\epsilon}-1 \right)\right]^{-1} \\ &\Rightarrow \frac{1}{x} < -\ln \left(\frac{2}{\epsilon}-1 \right) \\ &\Rightarrow - \frac{1}{x} > \ln \left(\frac{2}{\epsilon}-1 \right) \\ &\Rightarrow e^{- \frac{1}{x}} > \frac{2}{\epsilon}-1 \\ &\Rightarrow 1+e^{- \frac{1}{x}}>\frac{2}{\epsilon} \\ &\Rightarrow \frac{2}{1+e^{- \frac{1}{x}}}<\epsilon \\ &\Rightarrow \left|\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}-(-1)\right|=\frac{2e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}<\epsilon \end{align} This shows that $\lim_{x \rightarrow 0-}\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)=-1$. By taking the same $\delta$ it can be shown that $\lim_{x \rightarrow 0+}\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)=1$. Since the two limits aren't equal $\lim_{x \rightarrow 0}\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)$ doesn't exist.