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Evaluate the limit : $$\lim_{x\to 0^+} \frac{e^{-1/{x^2}}+2x}{e^{3x}\sin x}$$ The answer given is 2.

My Attempt:

As usual if we try to apply L'Hopital Rule $$\displaystyle\lim_{x\to 0^+} \frac{e^{-1/{x^2}}\left(\dfrac{2}{x^3}\right)+2}{3e^{3x}\sin x+e^{3x}\cos x}$$ But now it is no longer of the form $\frac{0}{0}$

So we would need to evaluate this separately: $\lim_{x\to 0^+}e^{-1/{x^2}}\left(\dfrac{1}{x^3}\right)$

Persistent use of the L-H rule doesn't seem to be giving any value

whoisit
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Maverick
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    That final limit and similar variants are well known and are equivalent to the fact that $\lim_{x\to\infty} x^r e^{-x}=0$ for all $r$ if that is something you have learned in class. It is more fundamental than LH – whpowell96 May 20 '25 at 01:31
  • I got your point. If I put $x=1/y$ then the limit i am trying to evaluate separately becomes $\lim_{y\to\infty}\frac{y^3}{e^{y^2}}$. On using LH rule twice I get the the limit to be $0$. Thanks – Maverick May 20 '25 at 01:51
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    Is this from this year’s JEE Advanced paper? – Integreek May 20 '25 at 11:04
  • Yes. It was a part of some other question – Maverick May 20 '25 at 15:04

2 Answers2

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I would just divide the numerator and denominator by $x$ without L'Hopital.

So you get $$\frac{\frac1xe^{-1/x^2}+2}{e^{3x}\frac{\sin x}x}$$

All of the terms are now converging other than possibly $\frac1xe^{-1/x^2}.$

The limit of that can be gotten by substituting $y=1/x,$ and seeking $$\lim_{y\to+\infty}ye^{-y^2}.$$

But $e^{y^2}>y^2$ so $$ye^{-y^2}<\frac{y}{y^2},$$ so that last limit is $0.$

You can do this trick with your remaining limit. For any $y>0,$ $$e^{y^2}>\frac{y^4}{2}$$ so $$0<y^3e^{-y^2}<\frac{2y^3}{y^4}\to0.$$

Thomas Andrews
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First, as $x\to0$, $$e^{3x}\sin x\sim x$$ hence $$\lim_{x\to0^+}\frac{e^{-1/x^2}+2x}{e^{3x}\sin x}=\lim_{x\to0^+}\frac{e^{-1/x^2}}x+2.$$ Then, $$\lim_{x\to0^+}\frac{e^{-1/x^2}}x=\lim_{y\to\infty}y^{1/2}e^{-y}=0,$$ as a special case of the (should be) well-known $$\lim_{y\to\infty}y^re^{-y}=0$$for all $r\in\Bbb R$ (see for instance How to find the value of $e^{-x} x^n$ at x = $\infty$? or What is the limit $x\to+\infty$ of $x^n \cdot e^{-x}$?).

Anne Bauval
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