<s>[INST] How to compute $\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use L'Hospital's rule.)
$$\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$$
 [/INST] Let's use the elementary techniques to solve the simple limit
\begin{align}
L &= \lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{4 - (1 + \cos x)^{2}}\notag\\
&= \lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{4 - 4\cos^{4}(x/2)}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{1 - \cos^{4}(x/2)}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{1 - \cos(x/2)}{1 - \cos^{4}(x/2)}\cdot\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{1 - \cos(x/2)}\notag\\
&= \frac{1}{4}\lim_{t \to 1}\frac{t - 1}{t^{4} - 1}\cdot\lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{1 - \cos(x/2)}\text{ (putting }t = \cos(x/2))\notag\\
&= \frac{1}{16}\lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{1 - \cos(x/2)}\notag\\
&= \frac{1}{16}\lim_{x \to 0}\frac{(x/2)^{2}}{1 - \cos(x/2)}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{(x/2)^{2}}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\sqrt{3e^{x} + e^{3x^{2}}} - 2}{x}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{(3e^{x} + e^{3x^{2}}) - 4}{x(\sqrt{3e^{x} + e^{3x^{2}}} + 2)}\notag\\
&= \frac{1}{8}\lim_{x \to 0}\frac{(3e^{x} + e^{3x^{2}}) - 4}{x}\notag\\
&= \frac{1}{8}\left(3\lim_{x \to 0}\frac{e^{x} - 1}{x} + \lim_{x \to 0}\frac{e^{3x^{2}} - 1}{3x^{2}}\cdot 3x\right)\notag\\
&= \frac{1}{8}(3\cdot 1 + 1\cdot 3\cdot 0)\notag\\
&= \frac{3}{8}\notag
\end{align}
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