Limit of some integral

Question

My question is how to find: $\displaystyle \lim_{n \rightarrow \infty} \int\limits_0^n \frac {1}{n+n^2\sin(xn^{-2})} dx $? I've tried with a dominated convergence theorem, but it didn't work. Now, I how absolutely no idea what I can do to solve it. Please, help me.

Answer 1

Let $t=\dfrac{x}n$. We then get $$I_n = \int_0^1 \dfrac{dt}{1+n \sin(t/n)}$$ Now use the fact that $\sin(t/n) > \dfrac2{\pi} \cdot \dfrac{t}n$ for $t \in [0,1]$. Hence, we get that $$1 + n \sin(t/n) > 1+\dfrac{2t}{\pi}$$ Hence, $$I_n < \int_0^1 \dfrac{dt}{1+\dfrac{2t}{\pi}} \text{, which is clearly bounded}$$ Hence, $$\lim_{n \to \infty}\int_0^1 \dfrac{dt}{1+n \sin(t/n)} = \int_0^1 \lim_{n \to \infty}\dfrac{dt}{1+n \sin(t/n)} = \int_0^1 \dfrac{dt}{1+t} = \log(2)$$

Answer 2

If you change the variable with $u=x/n^2$, you get $$ I_n=\int_0^{1/n}\frac{du}{\sin u+\frac{1}{n}}. $$

Now pick $\epsilon>0$.

For $n$ large enough $$ (1-\epsilon)u\leq \sin u\leq u $$ for all $u$ in $[0,1/n]$.

So $$ \log 2=\int_0^{1/n}\frac{du}{u+\frac{1}{n}}\leq I_n\leq \int_0^{1/n}\frac{du}{(1-\epsilon)u+\frac{1}{n}}=\frac{\log(2-\epsilon)}{1-\epsilon} $$ for all $n$ large enough.

So $$ \log 2\leq \liminf I_n\leq\limsup I_n\leq \frac{\log(2-\epsilon)}{1-\epsilon}. $$

Letting $\epsilon$ tend to $0$, we get $$ \lim I_n=\log 2. $$

Answer 3

Did you try the substitution $y = {x \over n}$? That looks a lot easier to evaluate to me.