Question

Find the limit.
\[\lim_{x \rightarrow \infty}(\sqrt{x^2+x}-\sqrt{x^2-x})\]

Answer 1

multiply top and bottom by the conjugate of the top

Answer 2

that is what you have is over 1 already

Answer 3

\[\frac{\sqrt{x^2+x}-\sqrt{x^2-x}}{1} \cdot \frac{\sqrt{x^2+x}+\sqrt{x^2-x}}{\sqrt{x^2+x}+\sqrt{x^2-x}}\]

Answer 4

once you do that I bet you can decide what to divide top and bottom by
if not let me know

Answer 5

I got 1 as the numerator, is that correct?

Answer 6

when you multiplied?

Answer 7

numerator: x^2+x-x^2-x

Answer 8

oh you forgot to distribute

Answer 9

you have
(x^2+x)-(x^2-x)
right?

Answer 10

x^2+x-x^2+x

Answer 11

oh oops!

Answer 12

by if you had what you said that would actually be 0 not 1

Answer 13

but anyways it is neither 0 or 1 in our case
it is actually just 2x on top

Answer 14

Yes, I got 2x

Answer 15

\[\lim_{x \rightarrow \infty}\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}\]

Answer 16

do you know what you are going to choose to divide top and bottom by?

Answer 17

no...

Answer 18

ok notice if the radical wasn't there there the highest exponent on bottom is 2
so you would have said x^2
but we have a radical over the x^2
so you divide top and bottom by sqrt(x^2)

Answer 19

\[\lim_{x \rightarrow \infty}\frac{\frac{2x}{\sqrt{x^2}}}{\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}}+\sqrt{\frac{x^2}{x^2}-\frac{x}{x^2}}}\]

Answer 20

now since sqrt(x^2)=x if x>0 then you can replace the top sqrt(x^2) with x

if you had x was going to negative infinity you would replace sqrt(x^2) with -x instead

Answer 21

Okay...
I got this
\[\frac{ 2 }{ 1+1 }= 2/2=1\]

Answer 22

great work

Answer 23

Thank you :)

Answer 24

np :)