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.
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.
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