Conjugate Limit: $\frac{\sqrt{x}-1}{x-1}$

[walker242]

$\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$

Homework Statement

Calculate the limit of \lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}.

Homework Equations

As above.

The Attempt at a Solution

Have tried to multiplicate with the conjugate.

 

[cristo]

Have tried to multiplicate with the conjugate.

Ok, what did you get? Note that you must show your work in order to get help here.

 

[walker242]

\lim_{x\to 1} \frac{\sqrt{x}-1}{x-1} = \lim_{x\to 1} \frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1)}{\left(x-1\right)\left(\sqrt{x}+1\right)} = \lim_{x\to 1} \frac{x-1}{x\sqrt{x}+x-\sqrt{x}-1} = \lim_{x\to1}\frac{x-1}{\sqrt{x}\left(x-1\right)+x-1} = \lim_{x\to1}\frac{x-1}{(x-1)(\sqrt{x}+1)} = \frac{1}{2}

So in essence, disregard me, for I am retarded. :P

 

[HallsofIvy]

For a retarded person, remarkably good at limits!

 

[rrogers]

How about LHopital's rule?

 

[HallsofIvy]

Why? That's like using a sledgehammer to crack a walnut. Walker242's solutions is excellent- especially because it is his solution!

 

[rrogers]

Why?
His solution is very good.
So I have no overriding reason; but LHopital is more generic.
But I am into generic, versus tricky.