# How to calculate $\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$?

I 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?

• Take the limit of the log of crud instead of the limit of crud. Then exponentiate your answer. – B. Goddard Sep 13 '16 at 23:17
• You mean like taking the limit of $e^{\ln{y}}$ (where $y$ is the whole expression) so that the exponent would just multiply with the logarithm of the base? – Radiant Sep 13 '16 at 23:41
• Yes, but I don't think of it as an exponent, because I can't see the tiny font then. I think after you take the log, L'hospital will work. – B. Goddard Sep 13 '16 at 23:51
• @B.Goddard I don't think it works here. Looking at the expression after the exponent has been "taken down", it is pretty clear that the limit of the log will be $\infty$ (because of the $e^{0^2}-1$ in the denominator. But the original limit is finite. – Bobson Dugnutt Sep 14 '16 at 0:14
• @Lovsovs But the numerator is $\ln (1)$, also 0. It may not work. The examples that break L'hospital usually have that sort of square root thing. Because of the $e^{x^2}$ someone would have to do L'hospital twice to get a non-zero term on the bottom, and that ain't gonna be me. – B. Goddard Sep 14 '16 at 0:23

## What is the problem?

From first inspection:

• $\cos(\sin x) + \frac{x^2}{2} \to 1^-$
• $e^{x^2}-1 \to 0^+$
• $1+2x - \sqrt{ (1+2x)^2 - 2x^2 }\to 0^+$

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 $0$: $$\frac{ \log\left( \cos(\sin x) + \frac{x^2}{2} \right) }{ (e^{x^2}-1) (1+2x - \sqrt{1+4x+2x^2}) }$$

## Using series expansion

### Choosing a precision

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).

### Numerator

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: $$\log\left(1 + \frac{5x^4}{24}\right) \sim \frac{5x^4}{24} + \mathcal{o}(x^4)$$

### Denominator

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: $$x^4 \left( 1 + \frac{x^2}{2} \right) \left( 1 +\mathcal{O}(x) \right)$$

## Final limit

Putting all this together, we finally obtain the desired limit: $$\exp\left[\frac{5}{24 \left(1+\frac{x^2}{2}\right) \left(1+\mathcal{O}(x)\right)}\right] \to e^\frac{5}{24}$$

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.

• Very nice moves :) I think once you get to $\big( \cos(\sin x) + x^2/2 - 1 \big) / x^4$ you can conclude easily with a development (see the numerator section in my answer). I find the gymnastic with the change of variables quite tedious after that step. – Sheljohn Sep 14 '16 at 1:33
• @Sh3ljohn: I prefer another approach. See my updated answer. I prefer to defer the use of series expansions or L'Hospital until they are necessary and in case of series expansion I try to use the standard expansions of $e^{x}, \sin x, \cos x$ etc. I never try to calculate the Taylor series coefficients and use only those which can be used directly from memory. – Paramanand Singh Sep 14 '16 at 1:42

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 $$\frac{1}{6}+\frac{1}{24}=\frac{5}{24}.$$

• Apart from this being a nice answer, it added "codicil" to my vocabulary. +1 – Paramanand Singh Sep 18 '16 at 14:01