Please show work

Attachment 27958

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- Apr 14th 2013, 06:19 PMleonhartFind Limit
Please show work

Attachment 27958 - Apr 14th 2013, 07:52 PMMINOANMANRe: Find Limit
factorize the denominator as (sqrx-1)(sqrx+1) then simplify and the limit is in front of you.........1/2

MINOAS - Apr 14th 2013, 09:45 PMEducatedRe: Find Limit
You can use l'hopital's rule, since you get 0/0 by direct substitution.

Which says:

$\displaystyle \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$

as long as the first limit is 0/0 or infinity/infinity

Therefore you get:

$\displaystyle \lim_{ x \to \1} \frac{ \sqrt{x} - 1}{x - 1} = \lim_{x\to\1} \frac{\frac{1}{2 \sqrt{x}}}{1} = \dfrac{1}{2}$ - Apr 15th 2013, 07:22 AMtopsquarkRe: Find Limit
As MINOANMAN said, but let's do a quick substitution to make it look a bit nicer. Let $\displaystyle x = y^2$. Then we have

$\displaystyle \lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} = \lim_{y \to 1} \frac{y - 1}{y^2 - 1} = \lim_{y \to 1} \frac{y - 1}{(y - 1)(y + 1)}$

etc.

-Dan - Apr 15th 2013, 07:51 AMBradynsRe: Find Limit
That is a rather elegant solution.