A limit to find without using l'Hôpital's rule - trigonometric functions.

Do you have some idea how to calculate limit without using L'Hospital rule? $$\lim_{x\to0} \frac{2^x-\cos\left(x\right)}{3^x-\operatorname{ch}\left(x\right)}$$

Thank you for help?

• What is "ch(x)"? Hyperbolic cosine? – DonAntonio Jan 9 '14 at 19:57
• @DonAntonio Some country uses the notations $ch(x)$ and $sh(x)$ and $th(x)$ and $tg(x)$ and $cotg(x)$ for $\cosh(x)$ and $\sinh(x)$ and $\tanh(x)$ and $\cot(x)$ respectively. – user63181 Jan 9 '14 at 20:06
• @SamiBenRomdhane Also, $ctg (x)$ for $cotg(x)$. – TZakrevskiy Jan 9 '14 at 21:25

$$\frac{2^x-\cos x}{3^x-\cosh x}\sim_0\frac{1+x\log2-1+o(x)}{1+x\log3-1+o(x)}\sim_0\frac{\log2}{\log3}$$
$$\frac{2^x-\cos x}{3^x-\cosh x}=\frac{\mathrm e^{x\log2}-\cos x}{\mathrm e^{x\log3}-\cos x}=\frac{1+x\log2+o(x)-(1+o(x))}{1+x\log3+o(x)-(1+o(x))}\to\frac{\log2}{\log3}$$