# 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.)

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

• Hint: for $x \rightarrow 0$ you have: <br/> $e^x = 1 + x + \dfrac{x^2}{2!} + o(x^3)$, $e^{3x^2} = 1 + 3x^2 + o(x^4)$, $\cos(x) = 1 - \dfrac{x^2}{2!} + o(x^4)$. Replace those expressions in the limit and do the math: you should find the result easily. – mrnld Jan 7 '16 at 21:41

Hint do you know about Maclaurin series?

You can write $$\lim _{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-\left(\cos x+1\right)^2}\right) = \lim _{x\to 0}\left(\frac{x\left(\sqrt{3 \cdot (1 + x + \frac{x^2}{2} + O(x^2)) +(1 + 3x^2 + O(3x^2))}-2\right)}{4-\left((1 - \frac{x^2}{2} + O(x^2))+1\right)^2}\right) = \lim _{x\to 0} \frac{\frac34 x^2 + \frac{63}{64} x^3 + O(x^3)}{2x^2-\frac{5}{12}x^4 + O(x^6)} = \lim _{x\to 0} \frac{\frac34 x^2 + O(x^2)}{2x^2 + O(x^2)} = \frac{\frac34}{2}\\ = \frac38$$

• what about the $\sqrt{}$ – user12 Jan 7 '16 at 21:57
• i've already counted in $\frac34 x^2 + \frac{63}{64} x^3 + O(x^3)$ – sirfoga Jan 7 '16 at 22:01
• so you have developed even $\sqrt{}$ with McLaurin? – user12 Jan 7 '16 at 22:03
• yup! the $\sqrt{}$ it' already in $\frac34 x^2 + \frac{63}{64} x^3 + O(x^3)$ – sirfoga Jan 7 '16 at 22:04
• i'm sorry i don't get the method for the $\sqrt{}$ can you please write it? – user12 Jan 7 '16 at 22:08

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}

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.

• $(1-\cos x)/x^2 = \sin^2 x/(1+\cos x)$ so no need for McClaurin – zhw. Jan 7 '16 at 21:59
• @zhw or perhaps even faster, $$1-\cos x=2\sin^2(x/2)$$ – Mark Viola Jan 7 '16 at 22:39
• I just wanted to show a possible approach related to the methods the OP might know. – Redundant Aunt Jan 8 '16 at 9:39