How to calculate $\lim_{n\to\infty}f(n)\sin(\frac{1}{n})$ where $f(x)=\int_{x}^{x^2}(1+\frac{1}{2t})^t\sin{\frac{1}{\sqrt{t}}}dt(x>0)$ 
$$f(x)=\int_{x}^{x^2}\left(1+\frac{1}{2t}\right)^t\sin\left(\frac{1}{\sqrt{t}}\right)dt\hspace{1cm}(x>0)$$
  try to find $$\lim_{n\to\infty}f(n)\sin\left(\frac{1}{n}\right)$$


I found this problem from a problem book and with a hint which tells me to apply L'Hospital's rule. 
But when $n\to\infty$ then $\frac{1}{\sin(1/n)}\to\infty$ and $f(n)\text{ seems}\to0$.
Would you help me with this problem? Best regards!
 A: It is best to find the derivative $f'(x) $ and it is evaluated using a combination of Fundamental Theorem of Calculus and Chain rule as $$f'(x) =\left(1+\frac{1}{2x^{2}}\right)^{x^{2}}\sin\left(\frac{1}{x}\right)\cdot 2x - \left(1+\frac{1}{2x}\right)^{x}\sin\left(\frac{1}{\sqrt{x}}\right)$$ and thus $f'(x) \to e^{1/2}\cdot 2-e^{1/2}\cdot 0=2\sqrt{e}$ as $x\to\infty$. It follows by L'Hospital's Rule that $$\lim_{x\to\infty} \frac{f(x)} {x} =2\sqrt{e}$$ and therefore $$\lim_{x\to\infty} f(x)\sin\left(\frac{1}{x}\right)=2\sqrt{e}$$ It follows that if $n$ is a positive integer then $$\lim_{n\to\infty}f(n) \sin\left(\frac{1}{n}\right)=2\sqrt{e}$$

In the above we have used two standard limits $$\lim_{x\to\infty} \left(1+\frac{t}{x}\right)^{x}=e^{t},\, \lim_{x\to\infty} x\sin\left(\frac{1}{x}\right)=1$$
A: $$f(x)=\int^{x^2}_x (1+\frac{1}{2t})^t \sin(\frac{1}{\sqrt{t}})dt$$
$$f(n)=\int^{n^2}_n (1+\frac{1}{2t})^t \sin(\frac{1}{\sqrt{t}})dt$$
$$\lim_{n\to \infty}f(n)\sin{\frac{1}{n}}=\lim_{n\to \infty}\sin{\frac{1}{n}}\int^{n^2}_n (1+\frac{1}{2t})^t \sin(\frac{1}{\sqrt{t}})dt$$
$$\lim_{n\to \infty}\sin{\frac{1}{n}}\int^{n^2}_n (1+\frac{1}{2t})^t \sin(\frac{1}{\sqrt{t}})dt$$
$$=\lim_{n\to \infty}\sin{\frac{1}{n}}\int^{n^2}_0 (1+\frac{1}{2t})^t \sin(\frac{1}{\sqrt{t}})dt-\sin{\frac{1}{n}}\int^{n}_0 (1+\frac{1}{2t})^t \sin(\frac{1}{\sqrt{t}})dt$$
$$=\lim_{n\to \infty}\frac{\int^{n^2}_0 (1+\frac{1}{2t})^t \sin(\frac{1}{\sqrt{t}})dt-\int^{n}_0 (1+\frac{1}{2t})^t \sin(\frac{1}{\sqrt{t}})dt}{\csc(\frac{1}{n})}$$
First show that this is in indeterminant form, then differentiate the top and bottom with L'Hopital.  You can do so with the Fundamental Theorem of calculus.
$$=\lim_{n\to \infty}n^2\sin^2(\frac{1}{n})\frac{2n(1+\frac{1}{2n^2})^{n^2} \sin(\frac{1}{n})-(1+\frac{1}{2n})^n \sin(\frac{1}{\sqrt{n}})}{\cos(\frac{1}{n})}$$
$$=\lim_{n\to \infty}n^2\sin^2(\frac{1}{n})\biggl(2n(1+\frac{1}{2n^2})^{n^2} \sin(\frac{1}{n})-(1+\frac{1}{2n})^n \sin(\frac{1}{\sqrt{n}})\biggr)$$
A: First prove that $f(n)\to\infty$ as $n\to\infty$ then apply L'Hopital rule
$$\lim_{n\to\infty}f(n)\sin\left(\frac{1}{n}\right)=\lim_{n\to\infty}\frac{f(n)}{\frac{1}{\sin\left(\frac{1}{n}\right)}}=\lim_{n\to\infty}\frac{f'(n)}{\left(\dfrac{1}{\sin\left(\frac{1}{n}\right)}\right)'}$$
$$f'(n)=2 \left(\frac{1}{2 n^2}+1\right)^{n^2} n \sin \left(\frac{1}{\sqrt{n^2}}\right)-\left(\frac{1}{2 n}+1\right)^n \sin \left(\frac{1}{\sqrt{n}}\right)$$
and
$$\left(\dfrac{1}{\sin\left(\frac{1}{n}\right)}\right)'=\frac{2 \cos \left(\frac{1}{n}\right)}{n^2 \left(1-\cos \left(\frac{2}{n}\right)\right)}$$
$$\lim_{n\to\infty} \frac{2 \left(\frac{1}{2 n^2}+1\right)^{n^2} n \sin \left(\frac{1}{\sqrt{n^2}}\right)-\left(\frac{1}{2 n}+1\right)^n \sin \left(\frac{1}{\sqrt{n}}\right)}{\frac{2 \cos \left(\frac{1}{n}\right)}{n^2 \left(\cos \left(1-\frac{2}{n}\right)\right)}}=$$
$$=\lim_{n\to\infty}\left[n^2 \sin \left(\frac{1}{n}\right) \left(2 \left(\frac{1}{2 n^2}+1\right)^{n^2} n \sin \left(\frac{1}{\sqrt{n^2}}\right)-\left(\frac{1}{2 n}+1\right)^n \sin \left(\frac{1}{\sqrt{n}}\right)\right) \tan \left(\frac{1}{n}\right)\right]$$
Now recalling that $$\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^t=e;\;\lim_{z\to 0}\frac{\sin z}{z}=1;\;\lim_{z\to 0}\frac{\tan z}{z}=1$$
you should arrive to the result $2 \sqrt{e}$
