Proving uniform convergence of $(1+\frac{x}{n})^n$ to $e^x$ on compact intervals in the real numbers My goal is to prove that if $b> a > 0$ are real numbers, then:
$\lim_{n \rightarrow \infty} \int_a^b (1 + x/n)^n e^{-x} dx = b-a$.
I think the best way to do this is to show that $(1+x/n)^n$ tends uniformly to $e^x$ on positive compact intervals. I know that it converges pointwise but I am struggling with the uniform proof. I have tried figuring out $\sup_{x \in [a,b]} \left|(1+\frac{x}{n})^n - e^x \right|$ by differentiating $(1+\frac{x}{n})^n - e^x$, but this is getting me nowhere. 
I would appreciate a nudge in the right direction about how to prove the convergence is uniform. Thank you!
 A: hint: Try to show $\left(1+\frac{x}{n}\right)^n\cdot e^{-x} \leq 1$,  and use $g(x) = 1$ as a dominating function, and use the DCT to conclude. And the above inequality is true because:  $\ln\left(1+\frac{x}{n}\right) \leq \frac{x}{n} $  is a well-known inequality.
A: $$\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
Substitute $n/x=m$
$$\lim_{m\to\infty}\left(1+\frac1m\right)^{mx}=\left(\lim_{m\to\infty}\left(1+\frac1m\right)^m\right)^x=e^x$$
If $x>0$, that is the limit.  If $x<0$, the limit is $\lim_{m\to\infty}$.
But you will find that when you go and prove this limit, all you get is $e^{\lim_{|m|\to\infty}1+\frac1m}$
So regardless of the direction you approach infinity, it evaluates to $e^x$
A: \begin{eqnarray}
e^x-(1+{x \over n})^n &=& \sum_{k>n} {1 \over k!} x^k  +\sum_{k=0}^n ({1 \over k!} - \binom{n}{k} {1 \over n^k}) x^k \\
&=& \sum_{k=0}^\infty a_{n,k} {1 \over k!} x^k \\
\end{eqnarray}
Where $a_{n,k} = \begin{cases} 1-(1(1-{1 \over n} ) \cdots (1 - { k-1 \over n})), & k \le n \\
1, & \text{otherwise}\end{cases}$. Note that $0 \le a_{n,k} \le 1$ for all $k, n$ and for any fixed $k$, then $\lim_n a_{n,k} = 0$.
Let $\epsilon>0$ and choose $N$ such that for $n \ge N$, $\sum_{k>n} {1 \over k!} b^k < {1 \over 2} \epsilon$.
Then
\begin{eqnarray}
|e^x-(1+{x \over n})^n| &\le& \sum_{k=0}^N a_{n,k} {1 \over k!} |x|^k + {1 \over 2} \epsilon \\
&\le& \sum_{k=0}^N a_{n,k} \max(1,b^N) + {1 \over 2} \epsilon \\
\end{eqnarray}
Now choose $N' \ge N$ such that $\sum_{k=0}^N a_{n,k} \le { 1\over 2 \max(1,b^N)} \epsilon$ for $n \ge N'$ (note the upper sum limit is fixed
at $N$) to get 
$|e^x-(1+{x \over n})^n| \le \epsilon $ for all $x \in [a,b]$.
A: Let $C=\max (|a|,|b|).$ For $k\geq 2$ let $2\leq n_k\in N$ such that $\forall x\in [a,b]\;(|x/n_k|<1/k).$ Then for $x\in [a,b]$  and $n\geq n_k$ we have $$|-x+(\log (1+x/n)^{n})|=$$ $$=|-x+x\sum_{j=1}^{\infty}(x/n)^j(-1)^{j+1}/j|=$$ $$=|x \sum_{j=2}^{\infty}(x/n)^{j-1}(-1)^{j+1}/j|\leq$$ $$\leq  C\sum_{j=2}^{\infty}k^{1-j}/j<C \sum_{j=2}^{\infty}k^{1-j}/2=C /2(k-1).$$ So $-x+\log (1+x/n)^n$ converges uniformly to $0$ on $[a,b].$ 
Since $\exp (x)$ is continuous, it is uniformly continuous on $[a,b],$ so $-\exp (x)+ \exp (\log (1+x/n)^n))$ also converges uniformly to $0$ on $[a,b].$
