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.

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