Question

√(x) - 1 ÷ x^2-1 limit ->1

Answer 1

\[\lim_{x \rightarrow 1}\frac{\sqrt{x}-1}{x^2-1} ?\]

Answer 2

yeah, thats it!

Answer 3

\[\lim_{x \rightarrow 1}\frac{\sqrt{x}-1}{(x-1)(x+1)}=\lim_{x \rightarrow 1}\frac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)(x+1)}\]

Answer 4

I applied Hopital's Rule and got 1/4 which is the same as myininaya.

Answer 5

\[=\lim_{x \rightarrow 1}\frac{1}{(\sqrt{x}+1)(x+1)}=\frac{1}{(\sqrt{1}+1)(1+1)}=\frac{1}{2(2)}=\frac{1}{4}\]

Answer 6

yes you can do l'hopital too! :)

Answer 7

\[(1/2x ^{-1/2})/(2x)\]
\[1/(4x ^{3/2})\]

Answer 8

i factored x-1 as (sqrt{x}-1)(sqrt{x}+1)

Answer 9

Hopital's only works since the top and bottom are 0 when eveluated at 1.

Answer 10

Nice factoring btw :)

Answer 11

thanks
nice "l'hopitaling" there

Answer 12

if you didn't like me factoring that
we could had done this alittle differently
\[\lim_{x \rightarrow 1}\frac{\sqrt{x}-1}{x^2-1}*\frac{\sqrt{x}+1}{\sqrt{x}+1}=\lim_{x \rightarrow 1}\frac{x-1}{(\sqrt{x}+1)(x^2-1)}\]
\[=\lim_{x \rightarrow 1}\frac{(x-1)}{(\sqrt{x}+1)(x-1)(x+1)}=\lim_{x \rightarrow 1}\frac{1}{(\sqrt{x}+1)(x+1)}\]