I am working with the Monotone, Dominant, and Bounded Convergence Theorems. I am attempting to evaluate the integral

lim$_{n\rightarrow\infty} \int_{0}^{1} \big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big) dx$

I have that $f_n(x) = \big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big)$ which is a continuous function on $[0,1]$ therefore it is Riemann Integrable.

$|f_n| = |\big(1+\displaystyle\frac{x}{n}\big)^{-n} \text{cos}\big(\frac{x}{n}\big)| \leq (1+0)^{-n} \cdot \text{cos}(0) = 0$

I am stuck on the rest. Now I have to show that $f_n\rightarrow$ to something almost everywhere.

  • $\begingroup$ I think it goes to $e^{-x}$. This isn't analysis or convergence theorems. This is just evaluating limits: $\lim_{n \to \infty} (1+\frac{x}{n})^{-n}\cos(\frac{x}{n})$ for $x \in (0,1)$ $\endgroup$ – mathworker21 Apr 3 '18 at 1:59
  • $\begingroup$ @mathworker21 the book gives an answer of $1+\frac{1}{e}$ $\endgroup$ – rover2 Apr 3 '18 at 2:01

It is a standard fact that $(1+\frac x n)^{n} \to e^{x}$ as $n \to \infty$. Hence $(1+\frac x n)^{-n} \to e^{-x}$ as $n \to \infty$. Since $\cos (\frac x n ) \to 1$ we get $f_n (x) \to e^{-x}$. To apply Dominated Convergence Theorem just note that $|f_n(x)| \leq 1$ (since $1+\frac x n \geq 1$ and $|\cos \frac x n| \leq 1$) and the constant function 1 is integrable. Hence the required limit is $\int_0^{1} e^{-x}dx=1-\frac 1 e$.

  • $\begingroup$ I searched for this theorem on the internet and I couldn't find it, so I wonder if you can please tell me if it is true or not, and send me a link to check it up if possible, and thanks in advance. The theorem is if $\lim_{x\to 0} x^{\alpha}f(x) =0$ and $\alpha >1$ then, $\int _0^1f(x) dx$ is convergent. Also another question please, how can we use this theorem for proving that $\int_0^R r \ln r dr$ is convergent? $\endgroup$ – Fareed Abi Farraj Sep 2 '19 at 13:49

Hint: Observe \begin{align} \left(1+\frac{x}{m} \right)^m \leq \left(1+\frac{x}{n} \right)^n \end{align} whenever $m\leq n$.

Additional Hint: Take $m=2$.

Here's the spoiler: Observe we have

\begin{align}\left( 1+\frac{x}{n}\right)^{-n} \leq \left(1+\frac{x}{2}\right)^{-2}\end{align}

for all $n\geq 2$. Hence we have that

\begin{align} \left|\left(1+\frac{x}{n} \right)^{-n}\cos\frac{x}{n} \right| \leq \left(1+\frac{x}{2}\right)^{-2}. \end{align}

Since $(1+x/2)^{-2}$ is integrable then by Dominated Convergence Theorem, we have that

\begin{align} \lim_{n\rightarrow \infty}\int^1_0\left(1+\frac{x}{n}\right)^{-n}\cos\frac{x}{n}\ dx = \int^1_0 e^{-x}\ dx = 1-e^{-1}.\end{align}


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