Question

I've been doing this problem for 2 days and I cant get the answer right. Can someone do it for me. I have like 3 alike this one. I just want to know the steps
Lim x->2 ((1/sqx+2)- 1/x)/x-2

Answer 1

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

Answer 2

like that?

Answer 3

it's that one

Answer 4

I know the answer should be 3/16 but I cant get to it

Answer 5

subtract

Answer 6

\[\frac{1}{\sqrt{x+2}}-\frac{1}{x}=\frac{x-\sqrt{x+2}}{x\sqrt{x+2}}\] might be a start

Answer 7

then maybe rationalize the numerator

Answer 8

yeah that will do it

Answer 9

\[\frac{x-\sqrt{x+2}}{x\sqrt{x+2}}\times \frac{x+\sqrt{x+2}}{x+\sqrt{x+2}}\]

Answer 10

leave the denominator in factored form , don't multiply it out
then numerator will have a factor of \(x-2\) in it
cancel with the \(x-2\) that i have avoided writing

Answer 11

there is where i get stuck

Answer 12

@n0z3kp0n3r

Using the difference of squares rule, you should get
\[\Large \left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right) = \left(x\right)^2-\left(\sqrt{x+2}\right)^2\]
\[\Large \left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right) = x^2-(x+2)\]
\[\Large \left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right) = x^2-x-2\]

Answer 13

So
\[\Large \frac{x-\sqrt{x+2}}{x\sqrt{x+2}}\]
turns into
\[\Large \frac{x^2-x-2}{\left(x\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)}\]

Answer 14

and I dont know what to do after that

Answer 15

I still have the (x-2) under that whole equetion

Answer 16

as a denominator

Answer 17

The numerator is actually (x-2)(x+1) so the (x-2) cancels

Answer 18

so we have this

\[\Large \frac{x^2-x-2}{(x-2)\left(x\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)}\]

Answer 19

yes it's easy afterwards when you've cnacelled the (x-2). The final form you can sub in x = 2 directly.

Answer 20

exactly as @mww states, the numerator x^2-x-2 factors to (x-2)(x+1) and the pesky x-2 term goes away. Once x-2 is gone from the denominator, you can use substitution to evaluate the limit

Answer 21

I think I got it. Thank you all. <3