Prove that $\displaystyle \lim_{k\rightarrow \infty}\int_0^k x^\alpha \left(1 -\frac{x}{k}\right)^k \, dx = \int_0^\infty x^\alpha e^{-x} \, dx$ for $\alpha > -1$.
I tried to use integration by part, binomial theorem, and power series expansion of $e^x$ but still fails to see how the equality holds.
It seems improbable that an instructor would assign this to students who are not expected to know that $\lim\limits_{k\,\to\,\infty} \left( 1 -\dfrac x k \right)^k = e^{-x},$ a fact usually stated in the standard first-year calculus course.
"Gary" posted a comment above that relies on the inequality that says for $u>0$ one has $\log_e u \le u-1.$ This can be seen by looking at the graph of $u\mapsto \log_e u.$ The tangent line to that graph at $u=1$ is the graph of $u\mapsto u-1.$
With $u = 1-\dfrac x k$ this inequality becomes $\log_e \left( 1 - \dfrac x k \right) \le -\dfrac x k,$ and hence $k\log_e\left( 1 - \dfrac x k \right) \le -x,$ so $\log_e\left(\left( 1 - \dfrac x k \right)^k \right) \le -x, \vphantom{\dfrac{\displaystyle\int}{}}$ and finally $\left( 1 - \dfrac x k \right)^k \le e^{-x}. $
Now notice that if $0\le x\le k$ then $1 - \dfrac x k \ge0.$ Thus $\left|\left( 1 - \dfrac x k \right)^k\right| \le e^{-x}.$
Let $\mathbf 1_{(0,k)}(x) = \begin{cases} 1 & \text{if } 0<x<k, \\ 0 & \text{if } x> k. \end{cases}$
So all of the functions $x^\alpha \cdot \mathbf 1_{(0,k)}(x) \cdot \left( 1 - \dfrac x k \right)^k$ are "dominated" by $x\mapsto x^\alpha e^{-x}$ on the interval $x>0.$
The dominated convergence theorem then says that if $\displaystyle \int_{(0,+\infty)} x^\alpha e^{-x}\,dx < +\infty$ then $$ \lim_{k\,\to\,\infty}\int\limits_{(0,+\infty)} x^\alpha \cdot \mathbf 1_{(0,k)} \cdot \left(1 - \frac x k \right)^k \,dx = \int\limits_{(0,+\infty)} \lim_{k\,\to\,\infty} x^\alpha \cdot \mathbf 1_{(0,k)} \cdot \left(1 - \frac x k \right)^k \,dx. $$
