Evaluate: $\lim_{n\to\infty} \int_{[0,\infty)} (1+x/n)^{-n}\sin(x/n)\,dx.$

Solution 1:

Hint: Using $$ (1+z)^n\ge 1+\binom{n}{2}z^2 $$ for $z>0$, one has $$ (1+\frac{x}{n})^n\ge\frac{n(n-1)}{2}\frac{x^2}{n^2}=1+\frac12(1-\frac1n)x^2\ge1+\frac14x^2$$ for $n\ge2$. Then define $$ f_n(x)=(1+\frac{x}{n})^{-n}\sin(\frac{x}{n}) $$ to estimate $f_n$ as $$ |f_n(x)|\le\frac{4}{4+x^2} $$ for $x\ge1$ and then one can use LDC.

Solution 2:

Here is another approach in case you didn't think to use the inequality provided by @xpaul. With $x \rightarrow x/n$ we see $$\int_0^{\infty}(1+x/n)^{-n}\sin(x/n)\mathrm{d}x=\int_0^{\infty}\frac{n\sin(x)}{(x+1)^n}\mathrm{d}x$$ Call this integral $a_n$. Verify yourself using $-1 \leq \sin(x) \leq 1$ that $|a_n|\leq 2$ for all $n>1$. Integrating by parts twice yields for $n>3$ $$a_n=\frac{n}{(n-1)(n-2)}\Bigg[1-\frac{a_{n-2}}{n-2}\Bigg]$$ Since $|a_{n-2}|\leq 2$ we have $$\frac{n}{(n-1)(n-2)}\Bigg[1-\frac{2}{n-2}\Bigg]\leq a_n\leq \frac{n}{(n-1)(n-2)}\Bigg[1+\frac{2}{n-2}\Bigg]$$ for all $n>3$. By the squeeze thereom $\{a_n\}_{n=2}^{\infty} \longrightarrow 0.$