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I am trying to find the following limit: $$\lim_{x\to 0} \frac{1}{x}\int_{x}^{2x}e^{-t^2}dt.$$

I tried to solve by using the Gaussian function, but in Gaussian function the usual limits of integration are from $-\infty$ to $\infty$ and with a certain upper bound $a$ the result of a Gaussian function ended up involving the erf function. I don't know how to find this limit or even if it exists or not by using these.

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    $\begingroup$ Hint: Use L'Hospital's Rule and the Fundamental Theorem of Calculus. $\endgroup$ Sep 16, 2015 at 15:24
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    $\begingroup$ Forget L'Hospital and... well, follow@Yves' hint, just posted. :-) $\endgroup$
    – Did
    Sep 16, 2015 at 15:29
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    $\begingroup$ And you don't even need the fundamental theorem of calculus. Just observe that $$\frac{1}{x} \int_x^{2x} f(t)\,dt$$ is the average of $f$ on the interval $[x,2x]$ (resp. $[2x,x]$ if $x < 0$). Thus if $f$ is continuous at $0$, it immediately follows that the limit is $f(0)$. $\endgroup$ Sep 16, 2015 at 15:45
  • $\begingroup$ @DanielFischer's remark is clearly optimal to solve this. $\endgroup$
    – Did
    Sep 16, 2015 at 15:55

5 Answers 5

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Let $g(x)$ be a primitive of $e^{-x^2}$. Then

$$\lim_{x\to 0}\frac{g(2x)-g(x)}x=\lim_{x\to 0}\left(2\frac{g(2x)-g(0)}{2x}-\frac{g(x)-g(0)}x\right)=2g'(0)-g'(0)=1.$$

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  • $\begingroup$ what exactly do you mean by primitive of $e^{-x^2}$. $\endgroup$
    – MrYouMath
    Sep 16, 2015 at 15:48
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    $\begingroup$ @MrYouMath A primitive of a function $f$ is a function $F$ with $F' = f$. You may be more accustomed to the modern neologism "antiderivative". $\endgroup$ Sep 16, 2015 at 15:49
  • $\begingroup$ Why is it called a primitive? Does this have its roots in algebra? $\endgroup$
    – MrYouMath
    Sep 16, 2015 at 15:52
  • $\begingroup$ I preferred primitive over antiderivative with the intent to avoid discussions about the existence of a closed formula representation, but that was a naïve idea. $\endgroup$
    – user65203
    Sep 16, 2015 at 16:14
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There is no need of De l'Hopital theorem: in a neighbourhood of zero, $e^{-t^2}=1-t^2+o(t^3)$, hence:

$$ \frac{1}{x}\int_{x}^{2x}e^{-t^2}\,dt = \frac{1}{x}\left(x-\frac{7}{3}x^3+o(x^4)\right) = 1 +O(x^2) $$ and the limit of the RHS as $x\to 0$ is clearly $\color{red}{1}$.

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Using the integral Mean Value Theorem, $$ \frac{1}{x}∫_x^{2x} f = \frac{∫_0^{2x} f - ∫_0^xf}{2x-x} = f(\xi(x))$$ for some $\xi(x) ∈ [x,2x]$. We then have $\xi(x) \xrightarrow[x→ 0]{} 0$, so by continuity of $f$, the result follows.

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WLOG, we assume that $x>0$. Then \begin{align*} \frac{1}{x}\int_x^{2x} e^{-t^2} dt &= \int_1^2 e^{-(ux)^2} du. \end{align*} Then \begin{align*} \lim_{x\rightarrow 0}\frac{1}{x}\int_x^{2x} e^{-t^2} dt &= \lim_{x\rightarrow 0}\int_1^2 e^{-(ux)^2} du\\ &= \int_1^2 \lim_{x\rightarrow 0} e^{-(ux)^2} du\\ &= 1. \end{align*}

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$$\frac{\int_{x}^{2x}e^{-t^2}dt}{x}=\frac{\int_{0}^{2x}e^{-t^2}dt+\int_{x}^{0}e^{-t^2}dt}{x}=\frac{\int_{0}^{2x}e^{-t^2}dt-\int_{0}^{x}e^{-t^2}dt}{x}$$

Now use L'Hospitals Rule and the fundamental Theorem of Calculus.

$$\lim_{x\to 0}\frac{\int_{x}^{2x}e^{-t^2}dt}{x}=\lim_{x\to 0}\frac{\int_{0}^{2x}e^{-t^2}dt-\int_{0}^{x}e^{-t^2}dt}{x}$$ $$=\lim_{x\to 0}\frac{2e^{-(2x)^2}-e^{-x^2}}{1}=1$$

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  • $\begingroup$ Corrected the little error in the calculation. $\endgroup$
    – MrYouMath
    Sep 16, 2015 at 15:51

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