I need some help with computing this limit:
$$\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$$
I'm guessing that I should do a Taylor expansion of $\cos(\sin x)$ in base, but what should I do with the stuff in the exponent?
The large expression in the exponent seems to be another technique for intimidating students and we can see from below evaluation of the limit that this large exponent just behaves like $1/x^{4}$ as $x \to 0$ (see from step marked $(A)$ to step marked $(B)$).
If $L$ is the desired limit then \begin{align} \log L &= \log\left\{\lim_{x \to 0}\left(\cos(\sin x) + \frac{x^{2}}{2}\right)^{1/(e^{x^{2}} - 1)(1 + 2x - \sqrt{1 + 4x + 2x^{2}})}\right\}\notag\\ &= \lim_{x \to 0}\log\left(\cos(\sin x) + \frac{x^{2}}{2}\right)^{1/(e^{x^{2}} - 1)(1 + 2x - \sqrt{1 + 4x + 2x^{2}})}\text{ (via continuity of log)}\notag\\ &= \lim_{x \to 0}\dfrac{\log\left(\cos(\sin x) + \dfrac{x^{2}}{2}\right)}{(e^{x^{2}} - 1)(1 + 2x - \sqrt{1 + 4x + 2x^{2}})}\tag{A}\\ &= \lim_{x \to 0}\frac{x^{2}}{e^{x^{2}} - 1}\cdot\dfrac{\log\left(\cos(\sin x) + \dfrac{x^{2}}{2}\right)}{x^{2}(1 + 2x - \sqrt{1 + 4x + 2x^{2}})}\notag\\ &= \lim_{x \to 0}\dfrac{\log\left(\cos(\sin x) + \dfrac{x^{2}}{2}\right)}{x^{2}(1 + 2x - \sqrt{1 + 4x + 2x^{2}})}\notag\\ &= \lim_{x \to 0}\dfrac{\log\left(\cos(\sin x) + \dfrac{x^{2}}{2}\right)}{x^{2}\{(1 + 2x)^{2} - (1 + 4x + 2x^{2})\}}\cdot\{(1 + 2x) + \sqrt{1 + 4x + 2x^{2}}\}\notag\\ &= \lim_{x \to 0}\dfrac{\log\left(\cos(\sin x) + \dfrac{x^{2}}{2}\right)}{x^{4}}\tag{B}\\ &= \lim_{x \to 0}\dfrac{\log\left(1 + \cos(\sin x) + \dfrac{x^{2}}{2} - 1\right)}{\cos(\sin x) + \dfrac{x^{2}}{2} - 1}\cdot\dfrac{\cos(\sin x) + \dfrac{x^{2}}{2} - 1}{x^{4}}\notag\\ &= \lim_{x \to 0}\dfrac{\cos(\sin x) + \dfrac{x^{2}}{2} - 1}{x^{4}}\tag{C}\\ &= \lim_{t \to 0}\frac{\cos(\sin 2t) + 2t^{2} - 1}{16t^{4}}\text{ (putting }x = 2t)\notag\\ &= \frac{1}{16}\lim_{t \to 0}\frac{2t^{2} - (1 - \cos(2\sin t\cos t))}{t^{4}}\notag\\ &= \frac{1}{16}\lim_{t \to 0}\frac{2t^{2} - 2\sin^{2}(\sin t\cos t)}{t^{4}}\notag\\ &= \frac{1}{8}\lim_{t \to 0}\frac{t + \sin(\sin t\cos t)}{t}\cdot\frac{t - \sin(\sin t\cos t)}{t^{3}}\notag\\ &= \frac{1}{8}\lim_{t \to 0}\left(1 + \frac{\sin(\sin t\cos t)}{\sin t\cos t}\cdot\frac{\sin t}{t}\cdot \cos t\right)\cdot\frac{t - \sin(\sin t\cos t)}{t^{3}}\notag\\ &= \frac{1}{4}\left(\lim_{t \to 0}\frac{t - \sin t\cos t}{t^{3}} + \frac{\sin t\cos t - \sin(\sin t\cos t)}{t^{3}}\right)\notag\\ &= \frac{1}{4}\left(\lim_{t \to 0}\frac{t - \sin t\cos t}{t^{3}} + \frac{\sin t\cos t - \sin(\sin t\cos t)}{(\sin t \cos t)^{3}}\cdot\frac{\sin^{3}t}{t^{3}}\cdot\cos^{3}t\right)\notag\\ &= \frac{1}{4}\left(\lim_{t \to 0}\frac{t - \sin t\cos t}{t^{3}} + \frac{1}{6}\right)\notag\\ &= \frac{1}{4}\left(\lim_{t \to 0}\frac{t - \sin t + \sin t (1 - \cos t)}{t^{3}} + \frac{1}{6}\right)\notag\\ &= \frac{1}{4}\left(\lim_{t \to 0}\frac{t - \sin t}{t^{3}} + \frac{\sin t}{t}\cdot\frac{1 - \cos t}{t^{2}} + \frac{1}{6}\right)\notag\\ &= \frac{1}{4}\left(\frac{1}{6} + 1\cdot\frac{1}{2} + \frac{1}{6}\right)\notag\\ &= \frac{5}{24}\notag \end{align} Hence $L = e^{5/24}$. Here we have freely used the standard limits $$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$ and the limits $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \frac{1}{6}$$ the first of which is an easy consequence of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$ and the second one can be easily established by using Taylor's series expansions or via L'Hospital's Rule.
Update: After step marked $(C)$ you may wish to use series expansions directly to get the answer quickly. Note that $$\cos (\sin x) = 1 - \frac{\sin^{2}x}{2} + \frac{\sin^{4}x}{24} + o(\sin^{4}x) = 1 - \frac{\sin^{2}x}{2} + \frac{\sin^{4}x}{24} + o(x^{4})$$ and hence $$\frac{\cos(\sin x) + x^{2}/2 - 1}{x^{4}} = \frac{x^{2} - \sin^{2}x}{2x^{4}} + \frac{\sin^{4}x}{24x^{4}} + o(1)$$ which is same as $$\frac{1}{2}\cdot\frac{x - \sin x}{x^{3}}\cdot\frac{x + \sin x}{x} + \frac{\sin^{4}x}{24x^{4}} + o(1)$$ and this tends to $$\frac{1}{2}\cdot\frac{1}{6}\cdot 2 + \frac{1}{24} = \frac{5}{24}$$ The step by step solution presented in first part of my answer is a bit lengthy because it is trying to avoid advanced tools (like Taylor and L'Hospital) as much as possible. When time is short (like an objective test) the second method is to be preferred.
This answer is a codicil to that of Paramanand Singh. I give here a different calculation of $$\lim\limits_{x\to 0} \frac{\cos(\sin x)-1+\frac{x^2}{2}}{x^4}$$
Frequent use of the replacement of $\sin y$ by $\frac{\sin y}{y}y$ is made.
We have $$ \frac{\cos(\sin x)-1+\frac{x^2}{2}}{x^4}= \frac{\cos(\sin x)-\cos x}{x^4} + \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}.$$ And we calculate each separately.
For the first we use the difference of cosines formula $\cos a-\cos b=2\sin \frac{a+b}{2}\sin \frac{b-a}{2}$
So we have $$\frac{\cos(\sin x)-\cos x}{x^4}=2\frac{\sin \frac{\sin x+x}{2}}{x}\frac{\sin \frac{x-\sin x}{2}}{x^3}$$ the standard replacement above reduces to the limit of $$\frac{1}{2}\frac{\sin x+x}{x}\frac{x-\sin x}{x^3}\rightarrow \frac{1}{6}.$$
As for the limit $\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$, many people would be willing to take this as granted but I give here a nice elementary derivation.
First $$\left(\frac{1-\cos x}{x^2}\right)^2=\frac{2-2\cos x-\sin^2 x}{x^4}\rightarrow \frac{1}{4}$$
Second $$\frac{x-\sin x}{x^3}\frac{\sin x}{x}=\frac{x\sin x-\sin^2 x}{x^4}\rightarrow \frac{1}{6}$$
Subtracting the first from the second gives
$$\frac{2\cos x-2 +x\sin x }{x^4}\rightarrow -\frac{1}{12} $$ we can rewrite this as
$$2\frac{\cos x-1 +\frac{x^2}{2} }{x^4}+\frac{\sin x -x}{x^3}\rightarrow -\frac{1}{12} $$ and this gives $$\frac{\cos x-1 +\frac{x^2}{2} }{x^4}\rightarrow \frac{1}{24} $$
Finally adding them together we get $$\frac{1}{6}+\frac{1}{24}=\frac{5}{24}.$$
From first inspection:
So we have a case of $1^\infty$, which is indeterminate, and therefore we are going to need series expansions to solve this. Let us develop the following exponent term near $0$: $$ \frac{ \log\left( \cos(\sin x) + \frac{x^2}{2} \right) }{ (e^{x^2}-1) (1+2x - \sqrt{1+4x+2x^2}) } $$
From rapid inspection we see that if we choose $\mathcal{O}(x^2)$, we will be left with $\cos(\sin x) + \frac{x^2}{2} \sim 1$, which will lead to $e^{\frac{\log 1}{0^+}}$ and it gets us nowhere. So we need at least $\mathcal{O}(x^3)$. Let us choose $\mathcal{O}(x^4)$ (you will see later why).
We have: \begin{align} \sin x &\sim x - \frac{x^3}{6} + \mathcal{o}(x^4) \\ \cos(\sin x) &\sim \cos\left( x - \frac{x^3}{6} \right) \sim 1 - \frac{\left( x - \frac{x^3}{6} \right)^2}{2} + \frac{\left( x - \frac{x^3}{6} \right)^4}{24} + \mathcal{o}(x^4) \end{align}
Note that the last term is important because it contains $x^4/24$ which is within the precision that we set. Most other terms vanish and we have after simplification: $$ \cos(\sin x) + \frac{x^2}{2} \sim 1 + \frac{5x^4}{24} + \mathcal{o}(x^4) $$
Now you can see why $\mathcal{O}(x^3)$ was not sufficient. Finally, we can develop the logarithm further: $$ \log\left(1 + \frac{5x^4}{24}\right) \sim \frac{5x^4}{24} + \mathcal{o}(x^4) $$
There are two factors in the denominator, one with an exponential, and one with a square-root, the relevant developments are: \begin{align} e^{x^2} - 1 &\sim x^2\left(1 + \frac{x^2}{2}\right) + \mathcal{o}(x^4) \\ [1+2x(2+x)]^{1/2} &\sim 1 + x(2+x) - \frac{x^2(2+x)^2}{2} + \mathcal{O}(x^3)\\ &\sim 1 + 2x - x^2 + \mathcal{O}(x^3) \end{align}
Note that we don't need to develop that last term to full precision because we only need to be able to factor $x^2$ to eliminate the factor $x^4$ in the numerator; once we do that, all other terms $\mathcal{o}(x^2)$ will vanish in the limit. (The actual development is $1+2x-x^2+2x^3-9x^4/2$).
This leads to the following denominator, after simplification: $$ x^4 \left( 1 + \frac{x^2}{2} \right) \left( 1 +\mathcal{O}(x) \right) $$
Putting all this together, we finally obtain the desired limit: $$ \exp\left[\frac{5}{24 \left(1+\frac{x^2}{2}\right) \left(1+\mathcal{O}(x)\right)}\right] \to e^\frac{5}{24} $$
