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I missed a couple of classes so I'm having trouble doing this (and other similar) excercises of homework:

$\lim\limits_{x\to0^+}\frac{\displaystyle\sqrt{x}-\displaystyle\int_0^\sqrt{x} \mathrm{e}^{-t^2}\,\mathrm{d}t}{\displaystyle\sqrt{x^3}}$

I am supposed to solve this limit using De L'Hospital rule, however I don't know how to take the derivative of the numerator.
Using the fundamental theorem of calculus I would know how to derivate $\displaystyle\int_0^x \mathrm{e}^{-t^2}\,\mathrm{d}t$, but I don't know what to do when I have a function of $x$ as a limit of the integral, even though I guess it's related to the fundamental theorem.

Is there a general rule to calculate the first derivative of $\displaystyle\int_a^{f(x)} f(t)\,\mathrm{d}t$ ?

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    $\begingroup$ Liebniz integral rule en.wikipedia.org/wiki/Differentiation_under_the_integral_sign $\endgroup$ – GTX OC May 13 '14 at 12:38
  • $\begingroup$ I like Marin's way best for simplicity. Otherwise we derivate and you should get $\frac{1/(2\sqrt{x})\cdot(1-e^{-x})}{3/2\cdot \sqrt{x}} = \frac{1-e^{-x}}{3x}$. Then derivate again and get $\frac{e^{-x}}{3} \rightarrow 1/3$ as $x \rightarrow 0$. This is in case you want verification of your answer. $\endgroup$ – Chris K May 17 '14 at 1:47
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You mean "calculate the derivative of $\int_a^{g(x)} f(t) dt$ with respect to $x$", in which case yes you can in this way:

Let $F(x) = \int_a^{x} f(t) dt$

Then $F'(x) = f(x)$ [by the fundamental theorem]

Thus $\frac{d}{dx}( F(g(x)) ) = (Fg)'(x) = F'(g(x)) g'(x) = f(g(x)) g'(x)$ [by chain rule]

It is good that you ask about the general question, but for this special case there is actually a faster way, because we can substitute $y=\sqrt{x}$, and $x \to 0^+$ is equivalent to $y \to 0^+$. Then the expression simplifies and we can apply the fundamental theorem directly.

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If you know enough about series and integrals to justify swapping of order between $\sum$ and $\int$ (or if you want to get a sense of what the limit is before trying to prove its value in a different way), you can do it this way (not the best way, but an entertaining one):

  • recall that for all $x\in\mathbb{R}$, $e^x \stackrel{\rm{}def}{=}\sum_{n=0}^\infty \frac{x^n}{n!}$;
  • write \begin{align*} \sqrt{x} - \int_0^{\sqrt{x}}e^{-t^2}dt &= \int_0^{\sqrt{x}}\left( 1-e^{-t^2} \right)dt \\ &= \int_0^{\sqrt{x}}\left( 1-\sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{n!} \right)dt \\ &= \int_0^{\sqrt{x}}\left( 1-1-\sum_{n=1}^\infty \frac{(-1)^n t^{2n}}{n!} \right)dt \\ &= -\int_0^{\sqrt{x}}\sum_{n=1}^\infty \frac{(-1)^n t^{2n}}{n!} dt \\ &= -\sum_{n=1}^\infty \frac{(-1)^n }{n!} \left( \int_0^{\sqrt{x}} t^{2n} dt\right)\tag{swap $(\dagger)$} \\ &= -\sum_{n=1}^\infty \frac{(-1)^n }{n!} \frac{x^{(2n+1)/2}}{2n+1} \\ &= \frac{x^{3/2}}{3} - \sum_{n=2}^\infty \frac{(-1)^n }{n!} \frac{x^{(2n+1)/2}}{2n+1} \end{align*} so that $$ \frac{\sqrt{x}-\int_0^\sqrt{x} {e}^{-t^2}dt}{\sqrt{x^3}} = \frac{1}{3} - \sum_{n=2}^\infty \frac{(-1)^n }{n!} \frac{x^{(2n+1)/2-3/2}}{2n+1} = \frac{1}{3} - \sum_{n=2}^\infty \frac{(-1)^n }{n!} \frac{x^{n-1}}{2n+1} = \frac{1}{3} + o(1) $$ (where the very last equality, $(\ddagger)$, also needs a brief justification). Hence, and modulo arguing$(\dagger)$ and $(\ddagger)$, this shows the limit is $1/3$.
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  • $\begingroup$ thanks for the answer, but I mostly studied by myself everything I know about series so I cannot understand your answer from the moment you swap the sum and the integral, I guess (hope?) it'll make sense to me in a few months! $\endgroup$ – Alessandro Codenotti May 13 '14 at 13:15
  • $\begingroup$ (Very) roughly, under certain assumptions on the series (here, it is easier to check them as this is a power series), you can do "whatever you want" (integration, differentiation, limits, etc) with infinite sums as if they were finite. $\endgroup$ – Clement C. May 13 '14 at 13:17
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\color{#00f}{\large\lim_{x \to 0^{+}}% {\displaystyle{\root{x} - \int_{0}^{\root{x}}\expo{-t^{2}}\,\dd t} \over \displaystyle{\root{x^{3}}}}} =\lim_{x \to 0^{+}}% {\displaystyle{x - \int_{0}^{x}\expo{-t^{2}}\,\dd t} \over \displaystyle{x^{3}}} =\lim_{x \to 0^{+}}% {\displaystyle{1 - \expo{-x^{2}}} \over \displaystyle{3x^{2}}} \\[3mm]&=\lim_{x \to 0^{+}}% {\displaystyle{2x\expo{-x^{2}}} \over \displaystyle{6x}} ={1 \over 3}\lim_{x \to 0^{+}}\expo{-x^{2}} =\color{#00f}{\large{1 \over 3}} \end{align}

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Other than Newton leibnitz theorem in differentiating the integral you have to use l' hospital's rule. i.e limit of f(x)/g(x) as x approaches to 0 is equal to limit of f'(x)/g'(x) as x approaches to 0

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