A limit

Evaluate the limit:

$$\lim_{n \to +\infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \,{\rm d}t$$

Solution:

 
We are using the following lemmas:

Lemma 1: $\displaystyle \lim_{n\rightarrow +\infty}\sum_{m=0}^{n}\frac{1}{\binom{n}{m}}=2$

Lemma 2: $\displaystyle \int_{0}^{1}x^m \left ( 1-x \right )^{n-m}\,{\rm d}x=\frac{1}{n+1}\cdot \frac{1}{\binom{n}{m}}$

Then for the limit we have successively:

$$ \begin{aligned}
\mathcal{J} &=\lim_{n\rightarrow +\infty}\int_{0}^{1}\frac{n+1}{2^{n+1}}\left [ \frac{\left ( t+1 \right )^{n+1}-\left ( 1-t \right )^{n+1}}{t} \right ]\,{\rm d}t \\
 &\overset{{\rm even \;\;\; to \;\; t }}{=\! =\! =\! =\! =\! =\! =\!}\frac{1}{2}\lim_{n\rightarrow +\infty}\int_{-1}^{1} \frac{n+1}{2^{n+1}}\left [ \frac{\left ( t+1 \right )^{n+1}-\left ( 1-t \right )^{n+1}}{t} \right ]\,{\rm d}t\\
 &\overset{t=2x-1}{=\! =\! =\! =\!}\lim_{n\rightarrow +\infty}\int_{0}^{1}(n+1)\left [ \frac{x^{n+1}-(1-x)^{n+1}}{2x-1} \right ]\,{\rm d}x \\
 &= \lim_{n\rightarrow +\infty}\int_{0}^{1}(n+1)(1-x)^n \left [ \frac{\left ( \frac{x}{1-x} \right )^{n+1}-1}{\frac{x}{1-x}-1} \right ]\,{\rm d}x \\
 &=\lim_{n\rightarrow +\infty}\int_0^1(n+1)\sum_{m=0}^{n}(1-x)^n \left ( \frac{x}{1-x} \right )^m\,{\rm d}x \\
 &=\lim_{n\rightarrow +\infty}\sum_{m=0}^{n}(n+1)\int_{0}^{1}x^m \left ( 1-x \right )^{n-m}\,{\rm d}x \\
 &=\lim_{n\rightarrow +\infty}\sum_{m=0}^{n}\frac{1}{\binom{n}{m}} \\
 &=2
\end{aligned}$$