I'm struggling to derive estimates for delta to prove the following limit,
$$\lim_{x\rightarrow 0}\frac{e^x-1}{e^{2x}-1}=\frac{1}{2}$$
What I've got: $$\forall \varepsilon >0, \exists \delta >0 \text{ such that } |x-0|<\delta \Rightarrow \left|\frac{e^x-1}{e^{2x}-1}-\frac{1}{2}\right|<\varepsilon.$$
Observe $$\frac{e^x-1}{e^{2x}-1}=\frac{1}{e^x+1}$$ and so we have
$$\left|\frac{1}{e^x+1}-\frac{1}{2}\right|=\left|\frac{e^x-1}{2(e^x+1)}\right|$$
Since the exponential function is increasing for all $x$ we have
$$|x|<\delta \\ \Rightarrow e^{-\delta}<e^x<e^\delta \\ \Rightarrow e^{-\delta}-1<e^x-1<e^\delta-1$$ and also $$\frac{e^\delta}{e^\delta+1}>\frac{1}{e^x+1}>\frac{1}{e^\delta+1}.$$
This is all I've gotten up to at the moment and clearly it becomes quite messy when I try to find an explicit form for $\delta(\varepsilon)$. Have I made a mistake somewhere?
Thanks
We want to make sure that $$\left|\frac{e^x-1}{2(e^x+1)}\right|\lt \epsilon.$$ Since $e^x\gt 0$ for all $x$, it is enough to make sure that $$|e^x-1|\lt \epsilon,$$ that is, $$1-\epsilon\lt e^x\lt 1+\epsilon.$$ Let us assume temporarily that $\epsilon\lt 1$. Then the inequality above will hold if $$\ln(1-\epsilon)\lt x\lt \ln(1+\epsilon).$$ Let $\delta=\min(\ln(2),\ln(1+\epsilon))$. If $|x|\lt \delta$, the desired inequality will hold.