Limit as n-> infinity with integral - Please check

[James LeBron]

Limit as n-> infinity with integral -- Please check

Homework Statement

Compute \lim_{n \to \infty} \int_{0}^{1} \frac{e^{x^4}}{n} dx.

Homework Equations

We can put the limit inside the integral as long as a function is continuous on a bounded interval, such as [0,1].

The Attempt at a Solution

I have a solution, and am just curious if I am using the right facts and/or rationale.

We consider a sequence of continuous functions f_{n} = \frac{e^{x^4}}{n} for x \in [0,1]. Since \lim_{n \to \infty} \frac{e^{x^4}}{n} = 0, then f_{n} converges pointwise to f(x) = 0.

Now we prove it converges uniformly. For a given \epsilon > 0, there exists N = \frac{e}{\epsilon} such that whenever n > N, we have |f_{n} - f| = |\frac{e^{x^4}}{n} - 0| < \epsilon.. We derived the value of N by knowing that since x \in [0,1], that \frac{e^{x^4}}{n} \le \frac{e}{n}.

Now we can just compute \int_{0}^{1} \lim_{n \to \infty} \frac{e^{x^4}}{n} dx.. This is just zero.

So what do you think?

 

[I like Serena]

Looks okay. Just as an observation.

Since n is constant within the integral, you can move it outside of the integral.
The integral itself is fully determined and will have some constant value.
Dividing this constant value by n will approach zero, so the limit is 0.

$$\lim_{n \to \infty} \int_{0}^{1} \frac{e^{x^4}}{n} dx = \lim_{n \to \infty} \frac 1 n \int_{0}^{1} e^{x^4} dx = \lim_{n \to \infty} \frac 1 n C = 0$$