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.)
$$\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$$
 A: 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}
A: Hint do you know about Maclaurin series?
You can write
$$\lim _{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-\left(\cos x+1\right)^2}\right) = \lim _{x\to 0}\left(\frac{x\left(\sqrt{3 \cdot (1 + x + \frac{x^2}{2} + O(x^2)) +(1 + 3x^2 + O(3x^2))}-2\right)}{4-\left((1 - \frac{x^2}{2} + O(x^2))+1\right)^2}\right) = \lim _{x\to 0} \frac{\frac34 x^2 + \frac{63}{64} x^3 + O(x^3)}{2x^2-\frac{5}{12}x^4 + O(x^6)} = \lim _{x\to 0} \frac{\frac34 x^2 + O(x^2)}{2x^2 + O(x^2)} = \frac{\frac34}{2}\\ = \frac38$$
A: To give another approach:
You can compute it by splitting it up:
$$
\lim_{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-(1+\cos(x))^2}\right)=\lim_{x\to 0}\left(\frac{\sqrt{3e^x+e^{3x^2}}-2}{x}\right)\lim_{x\to 0}\left(\frac{x^2}{4-(1+\cos(x))^2}\right)
$$
If you define $f(x)=\sqrt{3e^x+e^{3x^2}}$ and use $\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac{1}{2}$ (you can show this using maclaurin), this gives:
$$
\lim_{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-(1+\cos(x))^2}\right)=f'(0)\cdot \lim_{x\to 0}\left(\frac{1}{\left(\frac{1-\cos(x)}{x^2}\right)\left(3+\cos(x)\right)}\right)=\frac{1}{2}f'(0)
$$
It remains to calculate $f'(0)$ which is easy using the chain rule.
