The Purplemath Forums

by **trdert** on Wed Aug 05, 2009 8:47 pm

hi all,

i've got a couple problems i'm working on that require you to find the limit of a function with square roots. i'm having some issues with it. if anyone could lend a hand on the following problem it would be most appreciated:

lim_x->2 (sqrt(2x)-x)/(x-2)

i know i need to factor out x-2 but i don't know how to do that with the square root in the numerator.

thanks very much in advance!
tim

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by **stapel_eliz** on Wed Aug 05, 2009 9:04 pm

This is an interesting one. I think they're expecting you to multiply by the conjugate of the numerator to get:

. . . . . $\left(\frac{\sqrt{2x}\, -\, x}{x\, -\, 2}\right)\left(\frac{\sqrt{2x}\, +\, x}{\sqrt{2x}\, +\, x}\right)$

. . . . . $\frac{\left(\sqrt{2x}\right)^2\, -\, x^2}{(x\, -\, 2)\left(\sqrt{2x}\, +\, x\right)}$

. . . . . $\frac{2x\, -\, x^2}{-1(2\, -\, x)(\sqrt{2x}\, +\, x)}$

. . . . . $\frac{x(2\, -\, x)}{-1(2\, -\, x)(\sqrt{2x}\, +\, x)}$

Can you see where to go from there? :wink:

by **trdert** on Wed Aug 05, 2009 11:31 pm

fantastic, thanks! i'll review the whole multiplying by the conjugate thing also.

by **nona.m.nona** on Thu Aug 06, 2009 11:47 am

The conjugate thing comes up a lot on tests so, yeh, make sure you know it. :wink:

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finding limits with square roots in function

finding limits with square roots in function

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