Find the limit as x approaches infinity of (sqrt(1+3x^2))/x

  • Thread starter KevinL
  • Start date
  • #1
37
0

Homework Statement


Find the limit as x approaches infinity of (sqrt(1+3x^2))/x

The Attempt at a Solution


I tried using l'hopital's rule, but it gave me 3x/(sqrt(1+3x^2)) which doesnt help me at all.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31


use the fact that [itex]x= \sqrt{x^2}[/itex]
 
  • #3
jgens
Gold Member
1,581
50


Well, I can think of a couple of ways to do this one. Probably the simplest is to note that for arbitrarily large x, 3x^2 + 1 ~ 3x^2. Another way would be to multiply and divide the equation by x and then try to find the limit.
 
  • #4
726
1


Factor out a sqrt(x^2) from the numerator.
 
  • #5
37
0


Factor out a sqrt(x^2) from the numerator.
I dont believe you can just do that. Only if two things are multiplied beneath a radical can something be taken out.

Multiplying by x/x does not help either.

Hmm, perhaps you can say that as x gets arbitrarily large, its sqrt(3x^2)/x because the 1 becomes meaningless and thus the limit is sqrt(3)?
 
  • #6
jgens
Gold Member
1,581
50


Yes you can do that! However, multiplying and dividing by x certainly does help: Lim (x -> infinity) sqrt(1/x^2 + 3)/1.
 
  • #7
726
1


I dont believe you can just do that. Only if two things are multiplied beneath a radical can something be taken out.

Multiplying by x/x does not help either.

Hmm, perhaps you can say that as x gets arbitrarily large, its sqrt(3x^2)/x because the 1 becomes meaningless and thus the limit is sqrt(3)?


[tex]\sqrt(1+3x^2) = \sqrt(x^2(\frac{1}{x^2} + 3)) = \sqrt(x^2) \sqrt(\frac{1}{x^2} + 3)[/tex]
 
  • #8
37
0


Ah, clever. Thank you. Thats definitely how my prof would want me to do it.
 

Related Threads on Find the limit as x approaches infinity of (sqrt(1+3x^2))/x

Replies
4
Views
2K
Replies
4
Views
3K
Replies
13
Views
940
Replies
30
Views
7K
Replies
7
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
4
Views
2K
Replies
5
Views
1K
  • Last Post
Replies
10
Views
12K
  • Last Post
Replies
6
Views
11K
Top