Limit of an integral

Question

I'm not sure how to approach (no pun intended) the following limit:

$$\lim_{x \to 0^{+}} \sqrt{|\sin x - \tan x | } \int_{\cos x}^{1+ \sin x} e^y \, \, \mathrm{d}y$$

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I know that the indefinite integral of $e^y$ is just $e^y$, so we can rewrite the limit as

$$\lim_{x \to 0^{+}} \sqrt{|\sin x - \tan x |} \left(e^{1+\sin x} - e^{\cos x}\right)$$

We can also rewrite the difference of the $\sin $ and $\tan$ to get

$$\lim_{x \to 0^{+}} \sqrt{\left| \frac{\sin x \cos x - \sin x}{\cos x} \right|} \left(e^{1+\sin x} - e^{\cos x}\right)$$

Since $\sin 2x = 2 \sin x \cos x$, we have

$$\lim_{x \to 0^{+}} \sqrt{\left| \frac{\frac{1}{2}\sin 2x - \sin x}{\cos x} \right|} \left(e^{1+\sin x} - e^{\cos x}\right)$$

Maybe I could use the Taylor series for $\sin$ and $\cos$ to justify approximations like $\sin x \approx x$ for $x \ll 1$? I'm not sure ...

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EDIT: I think the answer is zero. I've written up my reasoning.

Answer 1

Taylor based approach seems to be a good way. Built at $x=0$, $$\sin(x)-\tan(x)=-\frac{x^3}{2}-\frac{x^5}{8}+O\left(x^7\right)$$ $$ \sqrt{|\sin x - \tan x |}=\frac{x^{3/2}}{\sqrt{2}}+\frac{x^{7/2}}{8 \sqrt{2}}+O\left(x^{9/2}\right)$$ $$e^{1+\sin x}=e+e x+\frac{e x^2}{2}+O\left(x^4\right)$$ $$e^{\cos x}=e-\frac{e x^2}{2}+O\left(x^4\right)$$ Combining all the pieces $$\sqrt{|\sin x - \tan x |} \left(e^{1+\sin x} - e^{\cos x}\right)=\frac{e x^{5/2}}{\sqrt{2}}+\frac{e x^{7/2}}{\sqrt{2}}+O\left(x^{9/2}\right)$$ From there, the limit of $0$ since the expression behaves as $x^{5/2}$.

If you plot on the same graph the expression and the approximation, you could be amazed to see how close they are. For example, you could check that, for $x=\frac {\pi}{6}$, the value of the expression is $\approx 0.585231$ while the approximation based on Taylor leads to $\approx 0.580961$ which is a quite good match even quite far away from $0$.

Answer 2

Let the original limit be $L$. We can rearrange to get

$$L = \lim_{x \to 0^{+}} \sqrt{\left| \frac{\sin x \cos x - \sin x}{\cos x} \right|} \left(e^{1+\sin x} - e^{\cos x}\right)$$

$$L = \left(\lim_{x \to 0^{+}} \sqrt{\left| \frac{\sin x (\cos x - 1)}{\cos x} \right|} \,\right) \cdot \left(\lim_{x \to 0^{+}} e^{1+\sin x} - e^{\cos x}\right)$$

Note that when $0 < x < 1$, $\cos x$ and $\sin x$ are both positive and less than $1$. In particular, $\cos x - 1$ is negative and $\frac{\sin x}{\cos x} > 0$, so

$$\left| \frac{\sin x (\cos x - 1)}{\cos x} \right| = \frac{(\sin x) (1- \cos x)}{\cos x}$$

We can substitute this in to get

$$L = \left(\lim_{x \to 0^{+}} \sqrt{\frac{(\sin x )(1- \cos x)}{\cos x}} \,\right) \cdot \left(\lim_{x \to 0^{+}} e^{1+\sin x} - e^{\cos x}\right)$$

The above step is where we use the fact that the limit is taken from the right (otherwise we cannot assume $0<x$). We can now simply plug in $x=0$. Since $\cos 0 = 1$ and $\sin 0 = 0$ we get

$$L = \left(\sqrt{\frac{(\sin 0 )(1- \cos 0)}{\cos 0}} \,\right) \cdot \left( e^{1+\sin 0} - e^{\cos 0}\right)$$

$$L = \sqrt{\frac{0 \cdot (1-0)}{1}}\cdot\left(e^{1+0}-e^1\right)$$

$$\boxed{L = 0}$$