$\lim_{n \to \infty } \int_0^n \left(1+\frac{(-x)^{27}}{n} \right) e^{-2x} dx$

Question

I'm practicing for the exam of Real Analysis (It is about Lebesgue integration in $\mathbb{R}^n$). I've encountered the following problem but I'm having some trouble solving it: $$\lim_{n \to \infty } \int_0^n \left(1+\frac{(-x)^{27}}{n} \right) e^{-2x} dx$$

My attempt: $$ \int_0^n \left(1+\frac{(-x)^{27}}{n} \right) e^{-2x} dx \leq \int_0^n \left(1+\frac{(-x)^{27}}{n} \right)dx = n - \frac{n^{27}}{28} \to -\infty \text{ as } n \to \infty $$ Therefore: $$\lim_{n \to \infty } \int_0^n \left(1+\frac{(-x)^{27}}{n} \right) e^{-2x} dx = -\infty$$

Is this right? Thanks for your help.

Answer 1

No, it is not. Don't "lose" the $e^{-2x}$ in your bound, it's the crucial factor to make things work.

Rewrite $$\int_0^n \left(1+\frac{(-x)^{27}}{n}\right)e^{-2x} dx = \int_{[0,\infty) }\underbrace{\left(1+\frac{(-x)^{27}}{n}\right)e^{-2x}\mathbf{1}_{[0,n]}(x)}_{f_n(x)} dx$$ and note that $$ \forall x_0\geq 0,\qquad f_n(x_0) \xrightarrow[n\to\infty]{} e^{-2x_0}\\\tag{pointwise convergence} $$ while $$ \forall n\geq 1, \forall x\geq 0, \qquad |f_n(x)| \leq \underbrace{(1+x^{27})e^{-2x}}_{g(x)} \tag{domination} $$ while $g$ is integrable on $[0,\infty)$: $ \int_{[0,\infty)} g(x)dx < \infty$.

Can you conclude via the Dominated Convergence Theorem?

Answer 2

If you multiply out the parens and split the integral into two, the first is

$$\int_0^n e^{-2x} \; dx \to \frac{1}{2}.$$

The other integral: Take the $\frac{1}{n}$ outside. Do integration by parts $27$ times. You get

$$\frac{1}{n}(\mbox{polynomial})e^{-2n} \to 0.$$

So the limit is $1/2.$