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Math Help - Evaluate limit (with integral)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Evaluate limit (with integral)?

    Evaluate:

    \lim_{x\rightarrow 0} \frac{1}{\sqrt{x}} \int_{2 - \sqrt{x}}^{2 + \sqrt{x}} e^{\sqrt{t^2}} dt
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  2. #2
    Senior Member Spec's Avatar
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    e^{\sqrt{t^2}}=e^{|t|}

    Since 2-\sqrt{x}\leq t \leq 2+\sqrt{x} and x is close to 0, t>0 for the interval we're integrating in.

    \lim_{x\to 0}\frac{1}{\sqrt{x}}\int_{2-\sqrt{x}}^{2+\sqrt{x}}e^{|t|} dt=\lim_{x\to 0}\frac{1}{\sqrt{x}}\int_{2-\sqrt{x}}^{2+\sqrt{x}}e^{t} dt= \lim_{x\to 0}\frac{1}{\sqrt{x}}(e^{2+\sqrt{x}}-e^{2-\sqrt{x}})=\{y=\sqrt{x},\ y\to 0, x\to 0 \}=\lim_{y\to 0}e^2\left(\frac{e^y-e^{-y}}{y}\right)= \lim_{y\to 0}2e^2\left(\frac{\sinh y}{y} \right)

    At this point we can use the Maclaurin series for \sinh x:

    \sinh x=x+\frac{x^3}{3!}+O(x^5)

    I'll leave the rest to you.
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  3. #3
    Super Member fardeen_gen's Avatar
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    The answer given in my text is simply 2e^2. Does that mean \lim_{y\rightarrow 0} \frac{\sinh y}{y} = 1 ?

    EDIT: Yes! I forgot the Maclaurin Series.
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  4. #4
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    Yes, you can see that when using the Maclaurin series since \frac{1}{x}\left(x+\frac{x^3}{3!}+O(x^5) \right)=1+\frac{x^2}{3!}+O(x^4) \to 1, x\to 0
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  5. #5
    Super Member fardeen_gen's Avatar
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    We can also do it like this :
    \lim_{y\rightarrow 0} e^2\left(\frac{e^y - e^{-y}}{y}\right)\ \ \left(\frac{0}{0}\ \mbox{form}\right)

    \mbox{Thus, by l'Hopital's rule,}

    Differentiating numerator and denominator,

    e^2\lim_{y\rightarrow 0} \left(\frac{e^y + e^{-y}}{1}\right)
    = 2e^2
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by fardeen_gen View Post
    Evaluate:

    \lim_{x\rightarrow 0} \frac{1}{\sqrt{x}} \int_{2 - \sqrt{x}}^{2 + \sqrt{x}} e^{\sqrt{t^2}} dt
    A slightly different solution :

    Let F be an antiderivative of t\mapsto \mathrm{e}^{\sqrt{t^2}}. We have

    \begin{aligned}<br />
\frac{1}{\sqrt{x}} \int_{2 - \sqrt{x}}^{2 + \sqrt{x}} e^{\sqrt{t^2}}\ \mathrm{d}t<br />
&=\frac{F(2+\sqrt{x})-F(2-\sqrt{x})}{\sqrt{x}}\\<br />
&=\frac{F(2+\sqrt{x})-F(2)}{\sqrt{x}}+\frac{F(2)-F(2-\sqrt{x})}{\sqrt{x}}\\<br />
&=\frac{F(2+\sqrt{x})-F(2)}{\sqrt{x}}+\frac{F(2-\sqrt{x})-F(2)}{-\sqrt{x}}<br />
\end{aligned}
    hence
    \begin{aligned}\lim_{x\to 0}\frac{1}{\sqrt{x}} \int_{2 - \sqrt{x}}^{2 + \sqrt{x}} e^{\sqrt{t^2}}\ \mathrm{d}t&= \lim_{h\to 0}\frac{F(2+h)-F(2)}{h}+\frac{F(2-h)-F(2)}{-h}\quad (\text{with}\ h=\sqrt{x})\\<br />
&=2F'(2)\\<br />
&=2\mathrm{e}^2<br />
\end{aligned}
    Last edited by flyingsquirrel; June 10th 2009 at 07:03 AM. Reason: + one step
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  7. #7
    Senior Member pankaj's Avatar
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    Leibnitz Rule:

    If F(x)=\int_{g(x)}^{h(x)}f(t)dt
    then
     <br />
F'(x)=f\left(h(x)\right)h'(x)-f\left(g(x)\right)g'(x)<br />

    = \lim_{x\to 0}\frac{\int_{2-\sqrt x}^{2+\sqrt x}e^{t^2}dt}{\sqrt x}

     <br />
=\lim_{x\to 0}\frac{e^{\sqrt{(2+\sqrt x})^2}\frac{1}{2\sqrt x}-e^{\sqrt{(2-\sqrt x})^2}\frac{1}{-2\sqrt x}}{\frac{1}{2\sqrt x}}<br />

    = e^2+e^2

    = 2e^2
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