PDA

View Full Version : $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$


walker242
Feb21-09, 06:04 AM
1. The problem statement, all variables and given/known data
Calculate the limit of \lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}.

2. Relevant equations
As above.

3. The attempt at a solution
Have tried to multiplicate with the conjugate.

cristo
Feb21-09, 06:06 AM
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
Feb21-09, 06:23 AM
\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
Feb21-09, 07:57 AM
For a retarded person, remarkably good at limits!

rrogers
Feb21-09, 09:27 AM
How about LHopital's rule?

HallsofIvy
Feb21-09, 09:38 AM
Why? That's like using a sledgehammer to crack a walnut. Walker242's solutions is excellent- especially because it is his solution!

rrogers
Feb21-09, 10:23 AM
Why?
His solution is very good.
So I have no overriding reason; but LHopital is more generic.
But I am into generic, versus tricky.