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The original is to calculate $$\lim_{n\rightarrow \infty } \int ^{\infty}_{0}\dfrac{n\sin \left(\frac {x}{n}\right)}{\left(1+\frac {x}{n}\right)^{n}}dx$$ or give a integral form.

I guess Lebesgue's dominated convergence theorem should work and the integral is $\int ^{\infty}_{0}\dfrac{x}{e^x}dx=1$. But I can't find the dominated function.Thanks!

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  • $\begingroup$ Hint: $\left(1+\frac{x}{n}\right)^n \ge \left(1+\frac{x}{3}\right)^3$ for $n \ge 3$. $\endgroup$ – achille hui Oct 10 '14 at 0:53
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The numerator is clearly bound by $n$, while the denominator is always smaller than $e^x$. The result follows.

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If you expand the integrand for large values of $n$, you get $$ \dfrac{n\sin \left(\frac {x}{n}\right)}{\left(1+\frac {x}{n}\right)^{n}}=e^{-x} x+\frac{e^{-x} x^3}{2 n}+\frac{e^{-x} x^3 \left(3 x^2-8 x-4\right)}{24 n^2}+\frac{e^{-x} x^5 \left(x^2-8 x+8\right)}{48 n^3}+\cdots$$ and you are just left with $$\int_0^{\infty}x^k e^{-x}dx=\Gamma (k+1)$$ So, $p$ being the number of terms used in the development, the value of the integral $I_p$ is $$I_0=1$$ $$I_1=\frac{n+3}{n}$$ $$I_2=\frac{n^2+3 n+6}{n^2}$$ $$I_3=\frac{n^3+3 n^2+6 n+5}{n^3}$$ $$I_4=\frac{n^4+3 n^3+6 n^2+5 n-33}{n^4}$$ $$I_5=\frac{n^5+3 n^4+6 n^3+5 n^2-33 n-266}{n^5}$$ $$I_6=\frac{n^6+3 n^5+6 n^4+5 n^3-33 n^2-266 n-1309}{n^6}$$

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