Limit as n-> infinity with integral - Please check

  • #1
Limit as n-> infinity with integral -- Please check

Homework Statement



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

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 [tex]f_{n} = \frac{e^{x^4}}{n}[/tex] for [tex]x \in [0,1][/tex]. Since [tex]\lim_{n \to \infty} \frac{e^{x^4}}{n} = 0[/tex], then [tex]f_{n}[/tex] converges pointwise to [tex]f(x) = 0[/tex].

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

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

So what do you think?
 

Answers and Replies

  • #2
I like Serena
Homework Helper
6,577
176


Looks okay. :smile:


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.

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

Related Threads on Limit as n-> infinity with integral - Please check

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
5
Views
3K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
8
Views
2K
Replies
5
Views
3K
Replies
5
Views
58K
Replies
1
Views
914
Replies
0
Views
1K
Top