Find the limit using delta epsilon definition. Evaluate $\lim_\limits{x \to 0}\ \dfrac{e^x-1}{e^{2x}-1}$ using $\delta - \varepsilon$ definition. 
Attempt: I claim that $\lim_\limits{x \to 0}\ \dfrac{e^x-1}{e^{2x}-1} = \dfrac 12$. $\forall \varepsilon >0, \exists\delta>0$ such that 
$$|\frac{e^x-1}{e^{2x}-1}-\frac12| = |\frac {2e^x-2-e^{2x}+1}{2e^{2x}-2}|=|\frac {2e^x-e^{2x}-1}{2e^{2x}-2}|\le |\cdot|<\varepsilon$$
I don't know how to proceed from here. I appreciate any hint. 

Will be easier to evaluate if you first factorise the denominator followed by the delta-epsilon evaluation
Hopefully you can proceed from there.
