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It's asked to solve this:

$$\lim_{x \to 0}\frac{\int_0^x(e^{2t}+t)^{1/t}dt}{x}$$

And I have no idea how to do it...

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    $\begingroup$ Have you tried applying L'Hopital's rule? $\endgroup$ – user170231 Jan 29 '15 at 15:22
  • $\begingroup$ So applying L'H I get $\lim_{x\to 0}(e^{2x}+x)^{1/x}$, so the answer is $\infty$? $\endgroup$ – João Pedro Jan 29 '15 at 15:32
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    $\begingroup$ Not quite. Substituting $x=0$ would give you $1^\infty$, which is actually an indeterminate form. $\endgroup$ – user170231 Jan 29 '15 at 15:34
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You can see this as the derivative of the function

$$F(x)=\int_0^x(e^{2t}+t)^{1/t}dt$$

At $x=0$ because $F(0)=0$

$$\lim_{x \to 0}\frac{\int_0^x(e^{2t}+t)^{1/t}dt-0}{x-0}=\lim_{x \to 0}\frac{F(x)-F(0)}{x-0}=F'(0)$$

Which by the fundamental theorem of calculus is equal to the function inside of the integral evaluated at $t=0$. But it happens that this value is not defined. So is needed to extend the function $(e^{2t}+t)^{\frac{1}{t}}$ continuously at $t=0$; this does not change the integral, and we can apply the FTC to the primitive of the extended function.

To compute that continuous extension, we need to compute a new limit which is the value of the function as $t \to 0$.

$$\lim_{t \to 0}(e^{2t}+t)^{1/t}=\exp{\lim_{t \to 0}\frac{\ln(e^{2t}+t)}{t}}$$ $$=\exp{\frac{d}{dt}\left(\ln{(e^{2t}+t)}\right)_{t=0}}=\exp{\left(\frac{2e^{2t}+1}{e^{2t}+t}\right)_{t=0}}=\exp{3}$$

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  • $\begingroup$ I don't follow the exp and ln step... I mean, you take ln on the fraction, and add exp to the left side, why? $\endgroup$ – João Pedro Jan 29 '15 at 16:16
  • $\begingroup$ When you have a limit $\lim_{x \to a} f(x)$ you can write it as $\lim_{x \to a}\exp (\ln f(x))=\exp \left(\lim_{x \to a}\ln f(x)\right)$ because $\exp$ is continous function. $\endgroup$ – rlartiga Jan 29 '15 at 16:19
  • $\begingroup$ exp(xxxx) notation means $e^{(xxxx)}$, right? $\endgroup$ – João Pedro Jan 29 '15 at 16:20
  • $\begingroup$ yes is the same. $\exp(x)=e^{x}$ $\endgroup$ – rlartiga Jan 29 '15 at 16:21
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    $\begingroup$ I think the formulation could be improved a bit. Where you say "is equal to the function inside of the integral" this refers to the value of that function at $0$. But it happens that this value is not defined. So what you do is try to extend the function $(e^{2t}+t)^{1/t}$ continuously at $t=0$; this does not change the integral, and you can apply the FTC to the primitive of the extended function. To compute that continuous extension, you need to compute a new limit (which then proceeds as indicated). $\endgroup$ – Marc van Leeuwen Jan 30 '15 at 8:17
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L'Hopital's rule is only step 1: $$\lim_{x \to 0}\frac{\int_0^x(e^{2t}+t)^{1/t}}{x} =\lim_{x \to 0}\frac{ \frac{d}{dx} \left(\int_0^x(e^{2t}+t)^{1/t}\right) }{1} = \lim_{x \to 0}(e^{2x}+x)^{1/x} $$ The next step is to expand $e^{2x}$: $$ \lim_{x \to 0}(e^{2x}+x)^{1/x} = \lim_{x \to 0}\left([1+2x+O(x^2)] + x\right)^{1/x} = \lim_{x \to 0}\left(1+3x\right)^{1/x} = e^3 $$

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  • $\begingroup$ What is this O? $\endgroup$ – João Pedro Jan 29 '15 at 15:41
  • $\begingroup$ Sorry man I'm just sick of this teacher. I'm graduating in engeneering and she is taking our skin out and her questions asks for more than needed, I'm exploiting this website. $\endgroup$ – João Pedro Jan 29 '15 at 15:43
  • $\begingroup$ For example, I'm in Calculus II and in Calculus I I never saw this f***** limit. $\endgroup$ – João Pedro Jan 29 '15 at 15:44
  • $\begingroup$ The expression $O(x^k)$ means "some function of $x$ that goes to zero sufficiently rapidly as x goes to zero that $\lim_{x\to 0}$ of that function divided by $x^k$ exists." $\endgroup$ – Mark Fischler Jan 29 '15 at 16:23
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    $\begingroup$ And in the context of this problem, what it means is that the terms beyond $2x$ in the expansion of $e^{2x}$ can safely be ignored. $\endgroup$ – Mark Fischler Jan 29 '15 at 16:25

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