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find $$\lim_{x \to \infty} e^x \sin (e^{-x^2})$$

As $x \to \infty$, we have $e^x \sin (e^{-x^2}) \to \infty \cdot 0$ , As we have an indeterminate form , we will use L'Hopital's rule.

I understood that the technique consists of rewriting the limit into either $\frac{0}{0}$ form or $\frac{\infty}{\infty}$ form. I choose the former.

So $$\lim_{x \rightarrow \infty} e^x \sin (e^{-x^2}) = \lim_{x \rightarrow \infty} \frac{\sin (e^{-x^2})}{\frac1{e^x}} = \lim_{x \rightarrow \infty} \frac{\left[\sin (e^{-x^2})\right]'}{\left[\frac{1}{e^x}\right]'} = \lim_{x \rightarrow \infty} \frac{\cos(e^{-x^2}) \cdot e^{-x^2} \cdot -2x}{-e^{-x}} $$

But I still find an indeterminate form. I tried as well by using the $\infty/\infty$ form. What would the best approach here to complete this?

Much appreciated.

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    $\begingroup$ You can always apply l'Hopital's rule again. $\endgroup$ – Michael McGovern Jul 15 '17 at 22:32
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    $\begingroup$ And yet another misdeed of L'H... $\endgroup$ – Did Jul 15 '17 at 22:33
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There are much easier approaches. Note that:

$$0<e^x\sin(e^{-x^2})<e^xe^{-x^2}=e^{x-x^2}$$

where we used $x>0\implies\sin(x)<x$. Now squeezing it is trivial, since $e^{x-x^2}\to e^{-\infty}=0$.

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$$\lim _{ x\rightarrow \infty } e^{ x }\sin \left( e^{ -x^{ 2 } } \right) =\lim _{ x\rightarrow \infty } \frac { \sin \left( e^{ -x^{ 2 } } \right) }{ { e }^{ -{ x }^{ 2 } } } \frac { { e }^{ x } }{ { e }^{ { x }^{ 2 } } } =\lim _{ x\rightarrow \infty }{ { e }^{ x-{ x }^{ 2 } } } =0$$

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  • $\begingroup$ could you elaborate how you go from $\lim _{ x\rightarrow \infty } \frac { \sin \left( e^{ -x^{ 2 } } \right) }{ { e }^{ -{ x }^{ 2 } } } \frac { { e }^{ x } }{ { e }^{ { x }^{ 2 } } }$ to $\lim _{ x\rightarrow \infty }{ { e }^{ x-{ x }^{ 2 } } }$ $\endgroup$ – gegu Jul 15 '17 at 23:21
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    $\begingroup$ @gegu,you know that $\lim _{ x\rightarrow 0 } \frac { \sin \left( x \right) }{ x } =1$ so $\lim _{ x\rightarrow \infty } \frac { \sin \left( e^{ -x^{ 2 } } \right) }{ { e }^{ -{ x }^{ 2 } } } =1$ $\endgroup$ – haqnatural Jul 15 '17 at 23:45
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Variant (simpler in my opinion):

Near $u=0$, $\sin u \sim u$, hence $$\mathrm e^x\sin \mathrm e^{-x^2}\sim_\infty e^x\mathrm e^{-x^2}=\mathrm e^{x-x^2}\to 0\quad\text{when}\quad x\to\infty.$$

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