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Find (with justification) $$ \lim_{n\to \infty} \int_0^n (1+x/n)^{-n}\log(2+\cos(x/n))\,dx $$

Ben Grossmann
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frusstu
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  • See this tutorial on how to format math properly. Note the changes that I have made for future reference. – Ben Grossmann Mar 30 '15 at 16:36
  • What have you tried here? The usual approach is to use dominated convergence. Does that apply here? Can you find a bounding function? – Ben Grossmann Mar 30 '15 at 16:37
  • Thank you for the edit. I am new at this. The function $$(1+x/n)^{-n} $$ is bounded by the constant function 1. However, this constant function is not integrable as n approaches infinity... – frusstu Mar 30 '15 at 16:53

2 Answers2

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Since $0<x<n$ it follows that $0<\dfrac{x}{n}<1$ and hence $\log(2+\cos(\dfrac{x}{n}))\leq\log3$.

Now $(1+\dfrac{x}{n})^{-n}=(1+\dfrac{x}{n})^{(n/x)(-x)}\leq2^{-x}$

So your dominating function is $3^{-x}\log 3$.

Now $(1+x/n)^{-n}\log(2+\cos(x/n))\to e^{-x}\log3$ so by Dominated Convergence Theorem,the integral converges to $\log3$.

Landon Carter
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Hint. You may observe that $$ \left(1+\frac x n \right)^{-n}\leq e^{-x}, \quad x\geq0, n \in \mathbb{N}^*, $$ and $$ \log (2+ \cos (x/n)) \leq \log (2+ |\cos (x/n)|) \leq \log (2+1)=\log 3 $$ thus $$ \mathbf{1}_{\large [0,n]}(x)\:\left(1+\frac x n \right)^{-n}\log (2+ \cos (x/n))\:\leq \mathbf{1}_{\large [0,\infty)}(x)\: e^{-x} \:\log 3 $$ and then use the dominated convergence theorem to conclude.

Olivier Oloa
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