Question

Limit as x approaches 1 of:
(sqrt(x) - 1)/(sqrt(x-1))
I can't use L'Hôpital's rule because we haven't learned it yet.

Answer 1

\[\huge \lim_{x \rightarrow 1}\frac{\sqrt x - 1}{\sqrt{x-1}}\]Multiplying top and bottom by the CONJUGATE of the top, gives us:
\[\huge \lim_{x \rightarrow 1}\frac{\sqrt x - 1}{\sqrt{x-1}}\left(\frac{\sqrt x + 1}{\sqrt x + 1}\right)\]

Answer 2

Remember what happens when you multiply conjugates? :D You get the difference of squares.

Answer 3

Hmmmm I'll bet this puts us on the right track :o hopefully..

Answer 4

Maybe not.. hmmmmm

Answer 5

Well, the denominator is still zero after 1 is plugged in.

Answer 6

@Zarkon to the rescue? :D hehe

Answer 7

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

\[\frac{x-1}{\sqrt{x-1}}=\sqrt{x-1}\]

Answer 8

\[\frac{\sqrt{x-1}}{\sqrt{x}+1}\to0\text{ as }x\to1\]

Answer 9

err... Why?

Answer 10

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

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

\[=\lim_{x \rightarrow 1}\frac{\sqrt{x-1}}{\sqrt x + 1}=\frac{0}{2}=0\]

Answer 11

wow, I'm blind. This is pretty obvious lol. Thanks for help!