How to evaluate $\lim _{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-\left(\cos x+1\right)^2}\right)$?

Question

I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Can you explain the method and the steps used?

$$\lim _{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-\left(\cos x+1\right)^2}\right)$$

Answer 1

You get in the denominator $$ 2^2-(1+\cos x)^2=(3+\cos x)·(1-\cos x)=(3+\cos x)·2 \sin^2(x/2) $$ so that you can reduce the limit to $$ \lim_{x\to 0}\frac{x^2}{(3+\cos x)·(1-\cos x)}·\lim_{x\to 0}\frac{\sqrt{3e^x+e^{3x^2}}−2}{x} \\ =\frac12·\lim_{x\to 0}\frac1{\sqrt{3e^x+e^{3x^2}}+2}·\lim_{x\to 0}\frac{3(e^x-1)+(e^{3x^2}-1)}{x} $$ which now has simple limits