Question

Find lim x goes to 0 for x square over 2-square root of x+4 and lim x goes to 2 for xsquare -x-2 over x square -4 and lim goes to 0 for sin square x over xsquare

Answer 1

wow thats a mouthful, can you post each limit separately using equation editor? it help clarify for people helping you
thx

Answer 2

I'm posting this from my phone m. Sorry

Answer 3

I really need help I only got two hours before mid trrm exam

Answer 4

\[\lim_{x \rightarrow 0} \frac{x^2}{2 \sqrt{x+4}}\] like this?

Answer 5

or subtract the square root?

Answer 6

You subtract it

Answer 7

\(\Large \lim \limits _{x \rightarrow 0} \frac{x^2}{2 -\sqrt{x+4}}\)
do you know how to rationalize the denominator ?

Answer 8

No

Answer 9

multiply and divide by conjugate of denominator, that is
\(2+\sqrt{x+4}\)
what do u get ?

Answer 10

The answer is 8 +4 square root of 6 over 6 is it correct ?

Answer 11

no, thats not correct. how did you get that ?

Answer 12

WE MILTIPLIED the conjugate .. I want the answers to anlazy it myself

Answer 13

\(\large \lim \limits _{x \rightarrow 0} \dfrac{x^2}{2 -\sqrt{x+4}} \dfrac{2 +\sqrt{x+4}}{2 +\sqrt{x+4}} = \dfrac{(2 -\sqrt{x+4)}x^2}{4- (x+4)} \\ \large =\dfrac{x^\cancel{2}( 2 -\sqrt{x+4})}{-\cancel x} = \dfrac{x (2 -\sqrt{x+4})}{-1}\)


now you can just plug in x =0 in this!
what do u get ?

Answer 14

Thanks man , you helped a lot ^^

Answer 15

welcome ^_^