Stuck trying to find the value of this limit using Taylor series.

Question

So I'm having trouble trying to find the value of the limit $$\lim_{x \to 0}\frac{(\sqrt{2}-\sqrt{1+\cos(x)})(e^x-e^{-x})}{\sin^3(x)}$$ using Taylor series (the problem explicits it). Well, this is obiously an $0/0$ indetermination so I first expanded $e^x$ and $e^{-x}$ up to order 3. I did the same thing on $\sin^3(x)$. The problem is that I don't know what to do with $\sqrt{1+\cos(x)}$. I tried to first expand it letting another variable that I called $t$ be $\cos(x)$ and the in every $t$ expanding cos(x) in a form that I only got terms of order 3, but I got the wrong answer so I thought that what I did was wrong. Could someone lend me a hand please?

Answer 1

Note that as $x\to 0$, $$\sqrt{2}-\sqrt{1+\cos(x)}=\frac{2-(1+\cos(x))}{\sqrt{2}+\sqrt{1+\cos(x)}} =\frac{x^2/2+o(x^2)}{2\sqrt{2}+o(1)}=\frac{x^2}{4\sqrt{2}}+o(x^2)$$ where we used the expansion $\cos(x)=1-x^2/2+o(x^2)$. Now it should be easy to get the job done.

Answer 2

Note that, near $0$,\begin{align}\sqrt{1+\cos(x)}&=\sqrt{1+\cos^2\left(\frac x2\right)-\sin^2\left(\frac x2\right)}\\&=\sqrt{2\cos^2\left(\frac x2\right)}\\&=\sqrt2\cos\left(\frac x2\right)\end{align}and that therefore\begin{align}\sqrt2-\sqrt{1+\cos(x)}&=\sqrt2\left(1-\cos\left(\frac x2\right)\right)\\&=\sqrt2\left(1-\cos^2\left(\frac x4\right)+\sin^2\left(\frac x4\right)\right)\\&=2\sqrt2\sin^2\left(\frac x4\right)\end{align}Can you take it from here?

Answer 3

$$\sqrt2-\sqrt{1+\cos x}=\sqrt2-\sqrt{2-\frac{x^2}2+o(x^2)}=\sqrt2-\sqrt2\left(1-\frac{x^2}8+o(x^2)\right)=\frac{x^2}{4\sqrt2}+o(x^2).$$ and $$e^x-e^{-x}=2x+o(x).$$

Then dividing these factors by $\sin^2x$ and $\sin x$ respectively, or equivalently by $x^2$ and $x$, the limit is

$$\frac1{2\sqrt2}.$$

Answer 4

The error in the calculation was: $\sqrt{1+t} = 1 + \frac{1}{2} t - \frac{1}{8} t^2 + \frac{1}{16} t^3 - \frac{5}{128} t^4 + \frac{7}{256} t^5 + \cdots$. But then, if you substitute $t = \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$, then $1, t, t^2, t^3, t^4, t^5, \ldots$ and so on all have a nonzero constant term, so to do the substitution you would need to sum an infinite series just to find the constant term, then another (more complicated) infinite series to find the $x^2$ coefficient, etc.

In order to correct this, let us instead set $t = \cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} - \cdots$. Then the function we want to expand is $\sqrt{2+t} = \sqrt{2} (1+t/2)^{1/2} = \sqrt{2} + \frac{\sqrt{2}}{4} t - \frac{\sqrt{2}}{32} t^2 + \frac{\sqrt{2}}{128} t^3 + \cdots$. Now, we see that $t^2, t^3, \ldots$ in terms of $x$ only contain $x^4$ and higher powers; so if we only want a third-order expansion, then we can ignore those terms. Thus, $$\sqrt{1 + \cos x} = \sqrt{2+t} = \sqrt{2} + \frac{\sqrt{2}}{4} \left(-\frac{x^2}{2} + O(x^4)\right) + O(x^4) = \sqrt{2} - \frac{\sqrt{2}}{8}x^2 + O(x^4).$$