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Find the lim x->4 of ((x^1/2)-2)/(x-4)
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Are we allowed to use L'Hospital rule?
yes.
i have the answer i just dont know how to get there.
Then you just need to differentiate the top and the bottom and then compute the limit.
yeah im completely stuck and dont understand it.
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$$\lim_{x\to 4} \frac{\sqrt{x}-2}{x-4}=\lim_{x\to 4}\frac{\frac{1}{2}x^{-1/2}}{1}=\frac{1}{4}$$
thanks!
Can also solve this one with conjugate of (sqrt x) -2 multiply top and bottom by (sqrt x) +2 that will eventually get you ((sqrt x)-2)^2 in the numerator which equals x-4 x-4's cancel giving you 1/(sqrt x) =2 in denominator : substitute limit 1/(2+2) = 1/4
That's correct! :D
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