finding the limit?

Question

lim (sqrt(9x^2+x)-3x) as x-->infinity

ok I know the answer already is 1/6
but I am unsure how they got that.
if you square it you get 9x^2-6xsqrt(9x^2+x)+9
if that is correct what do you do know? Square it again? to eliminate the sqrt once more?

Answer 1

It's best to deal first just with the square root term, noticing that it BEGINS with a perfect square. So write its argument as:

9^x^2 + x = 9 x^2 (1 + 1/(9x)) = (3x)^2 *(1 + 1/(9x)).

Now use the binomial theorem on sqrt (1 + 1/(9x)) :

Then 3x * sqrt (1 + 1/(9x)) = 3x * (1 + 1/2 * 1/(9x) + O(1/x^2))

= 3x (1 + 1/(18x) + ...) = 3x + 1/6 + ...

So lim (sqrt(9x^2+x)-3x) as x-->infinity = (3x + 1/6 + ...) - 3x

= 1/6 QED

Live long and prosper.

Answer 2

The most difficult of all indeterminate forms are those which are inf - inf, as is this one. There is no general method for handling these; do anything you can to change the form. Here, you could proceed as follows.

sqrt(9x^2 + x) - 3x = [sqrt(9x^2 + x) - 3x]*[sqrt(9x^2 + x) + 3x]/[sqrt(9x^2 + x) + 3x] = [9x^2 + x - 9x^2]/[sqrt(9x^2 + x) + 3x] = x/[sqrt(9x^2 + x) + 3x]. Now factor x^2 out from under the radical, and get x/[xsqrt(9 + 1x) + 3x], and cancel x from numerator and denominator. You have 1/[sqrt(9 + 1/x) + 3], which clearly has limit 1/6.