INFINITY Limit

This is a very challenging problem and is not in the same category as the other ones you've posted. I believe the answer is 1/6.

lim x-> inf (sqrt(9x^2+x) - 3x )=L
lim x-> inf sqrt(9x^2+x) = L + 3x
lim x-> inf 9x^2 + x = L^2 + 6Lx + 9x^2
lim x-> inf x = L^2 + 6Lx
lim x-> inf 1 = L^2/x + 6L

lim x-> inf L^2/x = 0
1 = 6L
L = 1/6

ok
Thank for your help. Our summer pre-calculus teacher is going over the basics of limits to give students a head start for the coming calculus 1 fall semester course.

Basics?
This is not from the basics of limits. I think a lot of people would have told you the limit was zero without looking at it closely.

Usually you can multiply both the numerator and the denominator by

(sqrt{9x^2 + x} + 3x)

without changing the value of

(sqrt{9x^2 + x} - 3x)=(sqrt{9x^2 + x} - 3x)/1

and then you can take the limit.
Answer is 1/6 as was shown above.

ok
Yes, in other words, satt_fs, you are saying that I could solve the question multiplying the original function by the conjugate?

Multiplying both the numerator and the denominator at the same time..

Well as far as I can see theres more hard algebra invovled than literal calculus. While it is not the basics of limits that kind of problem is fine for an intro into calculus since the real work is basically all the fundamentals of algebra that you know.

hey
What are the concepts introduced in a calculus 1 course? Does anyone know?

Very nice proof!