Question

lim x-->inf sqrt(x^2+18x) -x

How do you find the limit?

Answer 1

\[\large \lim_{x \rightarrow \infty}\sqrt{x^2+18x}-x\] Something to do with conjugates perhaps?

Answer 2

\[\large \lim_{x \rightarrow \infty }(\sqrt{x^2+18x}-x)\cdot \frac{(\sqrt{x^2+18x}+x)}{(\sqrt{x^2+18x}+x)}\]\[\large \lim_{x \rightarrow \infty } \frac{(x^{2}+18x-x^2)}{\sqrt{x^2 +18x}+x}\]\[\large \lim_{x \rightarrow \infty } \frac{18x}{\sqrt{x^2 +18x}+x}\]Divide by highest power in denominator. \[\large \lim_{x \rightarrow \infty } \frac{18x/x}{\sqrt{x^2/x^2 +18x/x^2}+x/x}\]\[\large \lim_{x \rightarrow \infty } \frac{18}{\sqrt{1 +0}+1}= \frac{18}{2} = 9\]

Answer 3

First turn it into a fraction so you can cancel stuff under a square root. Upon making it a fraction you divide by the highest power in the denominator which you simplify to find your answer.

Answer 4

And remember, depending on which limit you are taking (either from the left, or the right) you would either be dividing by +x or -x. And in dealing with the square root, you would be dividing by the highest power that's inside the square root,which is x^2.

Answer 5

How does 18x/x^2 go to zero?

Answer 6

Remember, you're not dividing by x^2 in the numerator. you're dividing by the highest power in the denominator, which is \(\large \sqrt{x^2} = \left|x\right| = +x \) sine you're tending to positive infinity.

Answer 7

since*

Answer 8

In the denominator, 18x/x^2 goes to zero, I don't understand how that works

Answer 9

Well, what happens when you divide x/x^2? What do you get?

Answer 10

1/x

Answer 11

the opposite of x^2/x

Answer 12

Ok. and so if we multiplied it by 18 we would have 18/x, correct?

Answer 13

as x approaches infinity, 18/x -> 0 because any constant divided by a large number like infinity will tend to 0.

Answer 14

Oh, ok, I get it now, the larger x gets, the smaller the number gets towards 0.

Answer 15

Thanks for your help!

Answer 16

No problem :) glad you understand this!

Answer 17

I'm trying to understand

Answer 18

Whats confusing you?

Answer 19

I was just saying in general, as it applies to calculus, that I'm trying to understand.

Answer 20

Well, it helps you determine asymptotes...helps determine the behavior of the function coming in from the left and right...etc.

Answer 21

I appreciate your help

Answer 22

no problem :)