Question

Limit as x approaches 1

(Square root of x+3)-2 / (x-1)

Help?

Answer 1

did you try to rationalize the numerator ?

Answer 2

multiply the numerator and denominator by
\(\sqrt{x+3}+2\)

Answer 3

Yes. That's where I got stuck on @hartnn

Answer 4

ok,then lets take the numerator

\( .\\ (\sqrt{x+3}-2)(\sqrt{x+3}+2) =.... ? \\~ use, (a+b)(a-b)=a^2-b^2\)

Answer 5

treat a as \(\sqrt{x+3} \)
and b=2

Answer 6

Will it be x-1? @hartnn

Answer 7

in the numerator, right ?yes!

Answer 8

so that gets cancelled with the x-1 in the denominator!
what remains ?

Answer 9

1/(square root of x+3)+2?

Answer 10

@hartnn

Answer 11

correct!
you can just plut x= 1 in that to get the limit! :)

Answer 12

Can you help me with another question @hartnn ?????

Answer 13

sure :)

Answer 14

Thank you !(:

Limit as x approaches 0 from the left side

Square root of -x

@hartnn

Answer 15

you can just plug in x =0 in that

Answer 16

It will be square root of -0..

Answer 17

-0 is 0 only
square root of 0 is 0 only

Answer 18

Okay. :) thanks

Answer 19

welcome ^_^

Answer 20

Wait @hartnn , help with my last one? :(

Answer 21

sure! :)

Answer 22

Okay. (:

Limit as x approaches 3 to the right

(Square root of x-3)/(square root of x-1)

Answer 23

you can again simply put x=3
because you're not getting any indeterminate form 0/0 or infinity/infinity here

Answer 24

So it would be correct if it's 0/(square root of 2) ??

Answer 25

which = 0 , yes!

Answer 26

Thank you so much ! Your a life saver :) @hartnn

Answer 27

you're most welcome ^_^