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

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


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 ...


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

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  • $\begingroup$ As a start, getting rid of the absolute value wouldn't hurt: $\tan x \ge \sin x$ for $x\in [0,\pi/2)$. $\endgroup$ – Henning Makholm Dec 14 '14 at 3:37
  • $\begingroup$ are you sure you have the product not the quotient?here both parts have zero as limit. you can conclude that the limit is zero. now, if it were the quotient, it would be another matter. $\endgroup$ – abel Dec 14 '14 at 12:42
  • $\begingroup$ Oh crap. You're right. It's actually $(\tan x - \sin x)^{-1/2}$ ... I had missed the negative sign in the exponent when I read it $\endgroup$ – Zubin Mukerjee Dec 14 '14 at 14:29
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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 $0$.

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  • $\begingroup$ I think this is incorrect. I'm not sure where you got the $0.58$ from, but I think the limit can be calculated without any Taylor series, and is equal to zero (see my answer). $\endgroup$ – Zubin Mukerjee Dec 14 '14 at 9:23
  • $\begingroup$ @ZubinMukerjee. The last equation I wrote shows that the limit is $0$ for sure. What I added was a comment telling that the approximation is quite good even far from $x=0$ (plot the two curves on the same graph). I shall clarify my answer. Thanks. $\endgroup$ – Claude Leibovici Dec 14 '14 at 9:26
  • $\begingroup$ Thank you. I am a bit uncertain about one step in my method ... do you think you could take a look? Specifically, I'm not sure if I'm allowed to plug in $x=0$ when I did - I think I can because the functions involved are continuous at $0$ but is there justification necessary? $\endgroup$ – Zubin Mukerjee Dec 14 '14 at 9:28
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    $\begingroup$ I must confess two things (but don't repeat !) : I am not good with limits ... except using Taylor series (I am in love with them for 55+ years). Let us see if you receive comments. Cheers :-) $\endgroup$ – Claude Leibovici Dec 14 '14 at 9:33
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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

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

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

$$\boxed{L = 0}$$

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