Question

find the limit if it exists. i think i have to do something with a conjugate on this one.
limg x->0 √2+x) -√2) / x

Answer 1

gonna take a shot at it. be back in a minute

Answer 2

\[\lim_{x\rightarrow0}\frac{\sqrt{2+x}-\sqrt2}{x}\]

Answer 3

yes just like that. still working on mine

Answer 4

multiply and divide by \(\sqrt{2+x}+\sqrt2\)

Answer 5

then use \((a-b)(a+b)=a^2-b^2\)

Answer 6

i can't remember for the life of me how to multiply the numerators.

Answer 7

im trying to use foil. by multiplying (√2x+x) -√2) (√2x+x)+√2), do i get rid of the radicals?

Answer 8

as i already mentioned use :\((a-b)(a+b)=a^2-b^2\)
here a= root(2+x), b =2

Answer 9

\[\lim_{x\rightarrow0}\frac{\sqrt{2+x}-\sqrt2}{x}=\lim_{x\rightarrow0}\frac{(\sqrt{2+x}-\sqrt2)(\sqrt{2+x}+\sqrt2)}{x(\sqrt{2+x}+\sqrt2)}=\lim_{x\rightarrow0}\frac{x}{x(\sqrt{2+x}+\sqrt2)}=\]\[\lim_{x\rightarrow0}\frac{1}{\sqrt{2+x}+\sqrt2}=\frac1{2\sqrt2}\]

Answer 10

i got the same denom as klime

Answer 11

I'm sorry. I have problems with internet.

Answer 12

yes,but did u get the numerator ?

Answer 13

nope. that's where i was getting messed up

Answer 14

gonna take a few to figure this out

Answer 15

i have to get going :( i'll be back in half an hour! thanks for your help so far. gonna finish this when i get back