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Find $\lim_{n \rightarrow \infty}$$\int_0^n(1 + \frac{-x}{n})^n\cos(\frac{x}{\sqrt{n}})e^{x/2}dx$

I want to use dominated convergence theorem obviously. However, not sure how to dominate it. Clearly $(1 + \frac{-x}{n})^n \rightarrow e^{-x}$.

Any suggestions? Thanks.

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    $\begingroup$ Try finding a $c > \frac{1}{2}$ such that $$\left( 1 - \frac{x}{n}\right)^n \leqslant e^{-cx}$$ for $0 \leqslant x \leqslant n$. $\endgroup$ – Daniel Fischer Jul 22 '14 at 18:51
  • $\begingroup$ Not seeing how to do this. $\endgroup$ – kingkongdonutguy Jul 22 '14 at 20:29
  • $\begingroup$ Hint: The exponential function is convex. $\endgroup$ – Daniel Fischer Jul 22 '14 at 20:38
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If we set $$ I_n = \int_{0}^{n}\left(1-\frac{x}{n}\right)^n\cos\frac{x}{\sqrt n}e^{x/2}dx$$ we obviously have: $$ I_n \leq \int_{0}^{n}\cos\frac{x}{\sqrt{n}}e^{-x/2}dx\leq \int_{0}^{\sqrt{n}}\cos\frac{x}{\sqrt n}e^{-x/2}dx+(n-\sqrt{n}) e^{-\sqrt{n}/2}\leq 2+o(1),$$ while: $$ \left|I_n-\int_{0}^{n}\cos\frac{x}{\sqrt{n}}e^{-x/2}dx\right|\leq 2o(1)+\int_{0}^{\sqrt{n}}\left (e^{-\xi}-\left(1-\frac{\xi}{n}\right)^n\right)e^{\xi/2}d\xi$$ is bounded by: $$2o(1)+C\cdot\int_{0}^{\sqrt{n}}\frac{\xi^2}{2n}\exp\left(\frac{\xi}{2}-\frac{(n-1)}{n}\xi\right)d\xi=o(1)+O(1/n) = o(1)$$ due to the inequality: $$ A\geq B\geq 0\quad\Rightarrow\quad A^n-B^n \leq n(A-B)A^{n-1}.$$ Hence we have: $$ I_n-2 = o(1)$$ and the limit is just $2$ by squeezing.

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  • $\begingroup$ How is the first inequality obvious? I don't see how $I_n≤\int_0^ncos\frac{x}{n^{0.5}}e^{−x/2}dx$ $\endgroup$ – kingkongdonutguy Jul 22 '14 at 22:18
  • $\begingroup$ $(1-\frac{x}{n})^n\leq e^{-x}$. $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 22:34
  • $\begingroup$ How do you prove that? I don't see it for some reason. $\endgroup$ – kingkongdonutguy Jul 22 '14 at 22:42
  • $\begingroup$ It is the Bernoulli inequality: for any $x$ such that $|x|\leq n$, $\left(1+\frac{x}{n}\right)^n\leq e^x.$ It just follows from the concavity of the logarithm. $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 22:49
  • $\begingroup$ All I am getting is that $(1 - \frac{x}{n})^n \geq 1 - x$. $\endgroup$ – kingkongdonutguy Jul 22 '14 at 23:33

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