# Solving the limit of integrals $\lim\limits_{q \to 0}\int_0^1{1\over{qx^3+1}} \, \operatorname{d}\!x$

how do I solve this one?

$$\lim_{q \to 0}\int_0^1{1\over{qx^3+1}} \, \operatorname{d}\!x$$

I tried substituting $t=qx^3+1$ which didn't work, and re-writing it as $1-{qx^3\over{qx^3+1}}$ and then substituting, but I didn't manage to get on.

Thanks in advance!

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if you substitute $t=q^{\frac{1}{3}}x$ you get $\int \frac{dt}{t^3+1}$, which is integrable, but the result is really messy –  Alex Apr 19 '13 at 16:15
I agree, I'll add the easy way –  julian fernandez Apr 19 '13 at 16:18

## 2 Answers

For any $q>0$, $$\frac{1}{q+1} = \int_0^1\frac{1}{q\cdot 1+1}\,\mathrm{d}x\leq\int_0^1\frac{1}{qx^3+1}\,\mathrm{d}x \leq \int_0^1\frac{1}{q\cdot 0+1}\,\mathrm{d}x = 1.$$ For any $q\in(-1,0)$, $$1 = \int_0^1\frac{1}{q\cdot 0+1}\,\mathrm{d}x\leq \int_0^1\frac{1}{qx^3+1}\,\mathrm{d}x\leq \int_0^1\frac{1}{q\cdot 1+1}\,\mathrm{d}x = \frac{1}{q+1}.$$

Thus, the limit for $q\to 0$ is $1$.

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Try substituting $t=q^{1/3}x$. Or, you can put the limit inside the definite integral first (the easy way)

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Is that 'allowed'? –  ohad Apr 19 '13 at 16:21
It is allowed, see en.wikipedia.org/wiki/Dominated_convergence_theorem, but I am not sure if you are supposed to assume you can (I mean, if it is homework). –  julian fernandez Apr 19 '13 at 16:23
Sadly we didn't prove this theorom in class so I still can't use it. Thanks anyway! –  ohad Apr 19 '13 at 16:31