Hard limit to solve involving exponential and square root How to find this limit:
$$\lim_{x \to 0}\frac{a(1-e^{-x})+b(e^x-1)}{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}.$$
Where $a$ and $b$ are integer constants. I've tried substitution, L'Hôpital's rule, rationalizing the denominator and all the techniques I knew about limits, but I can't find the answer.  

(assuming limit from the right). Use Taylor Series: $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\implies e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots$$
so that $$\lim_{x \to 0}\frac{a(1-e^{-x})+b(e^x-1)}{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}$$ $$=\lim_{x \to 0}\frac{a(x-\frac{x^2}{2}+\ldots)+b(x+\frac{x^2}{2}+\ldots)}{\sqrt{a(\frac{x^2}{2}-\frac{x^3}{3!}+\ldots)+b(\frac{x^2}{2}+\frac{x^3}{3!})}}\cdot\dfrac{\frac{1}{x}}{\frac{1}{x}}$$ $$= \lim_{x \to 0}\frac{a(1-\frac{x}{2}+\ldots)+b(1+\frac{x}{2}+\ldots)}{\sqrt{a(\frac{1}{2}-\frac{x}{3!}+\ldots)+b(\frac{1}{2}+\frac{x}{3!})}}=\dfrac{a+b}{\sqrt{\frac{1}{2}(a+b)}}=\boxed{\sqrt{2(a+b)}}$$
If it's the limit from the right, it's just the negative of this. Note that we used that $\frac{1}{x}=\frac{1}{\sqrt{x^2}}$ (principle).
Alternatively, we could have used L'Hopital just fine: $$I=\lim_{x \to 0}\frac{a(1-e^{-x})+b(e^x-1)}{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}$$
$$= \lim_{x \to 0}\frac{ae^{-x}+be^x}{\frac{1}{2}\left(a(e^{-x}+x-1)+b(e^x-x-1)\right)^{-\frac{1}{2}}\cdot \left(a(-e^{-x}+1)+b(e^x-1)\right)}=2(a+b)\lim_{n\to0}\frac{\sqrt{a(e^{-x}+x-1)+b(e^x-x-1)}}{a(1-e^{-x})+b(e^x-1)}=2(a+b)\frac{1}{I}\implies I^2=2(a+b)\implies \boxed{I=\sqrt{2(a+b)}}$$
