# Thread: Limit wit compliacated (looking) intergrals

1. ## Limit wit compliacated (looking) intergrals

$\lim_{\displaystyle x \to 0+} \frac{\displaystyle\int_{0}^{x}\left(\int_{0}^{t}\ sqrt{\displaystyle1+z^4}\mathrm{d}z\right) \mathrm{d}t}{\displaystyle\int_{0}^{x}\left(\int_{ 0}^{t}\sqrt{\displaystyle1+z^6}\mathrm{d}z\right) \mathrm{d}t}$

Any help would be appreciated!

2. You could try several approaches, I think. One is to use l"Hospital's rule. Another is to reverse the order of integration in both numerator and denominator. You should be able to get traction one way or the other. How does that strike you?

3. Thank you for your help.
I am afraid, I still can't solve the problem
I uderstand that $\lim_{\displaystyle x \to 0+}\displaystyle\int_{0}^{x}\left(\int_{0}^{t}\sqr t{\displaystyle1+z^4}\mathrm{d}z\right) \mathrm{d}t=\lim_{\displaystyle x \to 0+}\displaystyle\int_{0}^{x}\left(\int_{0}^{t}\sqr t{\displaystyle1+z^6}\mathrm{d}z\right) \mathrm{d}t=0$
so we can use L'Hospital's rule. But I don't know how to derive these expressions (with respect to z?).
So I need a little more help!
Thank you!

4. $F(x) = \int_0^x {\int_0^t {\sqrt {1 + z^4 } dz} dt} \; \Rightarrow \;F'(x) = \int_0^x {\sqrt {1 + z^4 } dz} \; \Rightarrow \;F''(x) = \sqrt {1 + x^4 }$

5. Originally Posted by doug
$\lim_{\displaystyle x \to 0+} \frac{\displaystyle\int_{0}^{x}\left(\int_{0}^{t}\ sqrt{\displaystyle1+z^4}\mathrm{d}z\right) \mathrm{d}t}{\displaystyle\int_{0}^{x}\left(\int_{ 0}^{t}\sqrt{\displaystyle1+z^6}\mathrm{d}z\right) \mathrm{d}t}$ Any help would be appreciated!
$\displaystyle \lim_{x \rightarrow 0+} \frac{\int_{0}^{x} \int_{0}^{t} \sqrt{1+z^{4}}\ dz\ dt}{\int_{0}^{x} \int_{0}^{t} \sqrt{1+z^{6}}\ dz\ dt} = \lim_{x \rightarrow 0+} \frac{\int_{0}^{x} \sqrt{1+z^{4}}\ dz} {\int_{0}^{x} \sqrt{1+z^{6}}\ dz}=$
$\displaystyle = \lim_{x \rightarrow 0+} \frac{\sqrt{1+x^{4}}}{\sqrt{1+x^{6}}} = 1$
$\chi$ $\sigma$