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I have the following limit to solve:

$$\lim_{x\to 0} \frac{\int_0^{x^2} (e^{t^2}-1)~dt}{\log(1+2x^6)}$$

The only thing I can do I guess is applying de l'Hôpital's rule and obtain:

  • Numerator: $$ (e^{x^4} - 1)\times 2x $$
  • Denominator: $$ \frac{12x^5}{1+2x^6} $$

What can I do now? Note that for the numerator I have used the theorem of integrals which says that the integral above (with $g(x)$ as upper bound and $h(dx)$ as lower bound) is:

$$ f(g(x))*g(x)' - f(h(x))*h(x)' $$

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    If you plot the function, the result should be infinite – 2017-01-15

1 Answers 1

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$\displaystyle \lim_{x\rightarrow 0}\frac{\int^{x^2}_{0}(e^{t^2}-1)dt}{\ln(1+2x^6)}$

applying de l'Hôpital's rule and $\displaystyle \lim_{y\rightarrow 0} \frac{e^y-1}{y} = 1$

$\displaystyle \lim_{x\rightarrow 0} \frac{(e^{x^4}-1)\cdot 2x}{\frac{12x^{5}}{1+2x^6}}=\lim_{x\rightarrow 0}2\frac{e^{x^4}-1}{x^4}\cdot \lim_{x\rightarrow 0} \frac{1+2x^6}{12} = 2\times \frac{1}{12}$

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    I got the exact same result but I don't thing it's correct, if you try with Wolfram you get +inf or -inf if you are at 0+ or 0- – 2017-01-15
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    @RaffaeleRossi Wolfram Alpha seems to give the same answer: http://www.wolframalpha.com/input/?i=lim_%7Bx+to+0%7D+(int_0%5E(x%5E2)+(e%5E(t%5E2)-1)+dt)%2Fln(1%2B2x%5E6) – 2017-01-15