I've made an attempt at the following question but am not sure if I'm heading in the right direction:
"Evaluate the limit of the following integral
$lim_{n \rightarrow \infty} \int_{[0,1]} (\frac{1 - sin^2 (x)}{n})^6 dm(x)$"
My attempt is as follows:
Let $f(x) = (\frac{1 - sin^2(x)}{n})^6$
Then $\frac{sin^2(x)}{n} \leq \frac{sin^2(x)}{n + 1}$ $\leftrightarrow$ $\frac{1 - sin^2(x)}{n} \geq \frac{1 - sin^2(x)}{n+1}$ $\leftrightarrow$ $(\frac{1 - sin^2(x)}{n})^6 \geq (\frac{1 - sin^2(x)}{n+1})^6$
I'm not sure if what I'm doing is correct and any help would be much appreciated, thanks.
Well $1-\sin(x)^2 = \cos(x)^2$ so we have that $| (\frac{1-\sin^2(x)}{n})^6| \leq \frac{1}{n^6}$. So by lebesgues dominated convergence we may pass the limit inside the integral, and we get: $$ \lim_{n\to \infty} \int_{[0,1]} (\frac{1-\sin^2(x)}{n})^6 dm(x) = \int_{[0,1]}0dm(x) = 0. $$