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anonymous
 one year ago
find lim as x approaches 4 of ((sqrt(3x+4)sqrt(4x))/(x^24x)
anonymous
 one year ago
find lim as x approaches 4 of ((sqrt(3x+4)sqrt(4x))/(x^24x)

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dan815
 one year ago
Best ResponseYou've already chosen the best response.2both the top and bottom are going to 0, so you can apply l'hopitals rule

dan815
 one year ago
Best ResponseYou've already chosen the best response.2differentiate the top and differentiate the bottom separetly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0duh I forgot that square roots are another form of power to the half

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439928493843:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2.....and a question the numerator isn't always 0 \(\sqrt{16}  \sqrt{16} = \pm4 \  \pm 4 \implies 8, 0 , 8\)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0i think this square root is the principal square root you could also do this without l'hospital you can rationalize the numerator and then which eventually leads to canceling the (x4) factor out on top and bottom and then you will be able to do direct substitution.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2thats a good point irish

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@myininaya thank you!! \(\large \frac{\sqrt{3x+4}\sqrt{4x}}{x^2  4x} .\frac{\sqrt{3x+4}+\sqrt{4x}}{\sqrt{3x+4}+\sqrt{4x}}\) \(= \frac{3x+44x}{x(x4)(\sqrt{3x+4}+\sqrt{4x})}\) \(= \frac{(x4)}{x(x4)(\sqrt{3x+4}+\sqrt{4x})}\) \(= \frac{1}{x(\sqrt{3x+4}+\sqrt{4x})}\) \(= \frac{1}{4(\sqrt{16}+\sqrt{16})}\) \(= \frac{1}{32}\) not familiar with this trick

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2like a conjugate, i guess, but having the nous to know that it works for the denominator too.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0here is a fun one and sorta similar one: \[\lim_{x \rightarrow 2}\frac{\sqrt{6x}2}{\sqrt{3x}1} \\ \lim_{x \rightarrow 2} \frac{\sqrt{6x}2}{\sqrt{3x}1} \cdot \frac{\sqrt{3x}+1}{\sqrt{3x}+1} \cdot \frac{\sqrt{6x}+2}{\sqrt{6x}+2} \\ \lim_{x \rightarrow 2}\frac{ 6x4}{3x1} \frac{\sqrt{ 3x}+1}{\sqrt{6x}+2} \\ \lim_{x \rightarrow 2} \frac{x+2}{x+2} \frac{\sqrt{3x}+1}{\sqrt{6x}+2} \\ \lim_{x \rightarrow 2} \frac{\sqrt{3x}+1}{\sqrt{6x}+2} \\ = \frac{\sqrt{32}+1}{\sqrt{62}+2}=\frac{2}{2+2}=\frac{1}{1+1}=\frac{1}{2}\] I always thought the two conjugate thing was really cute for some reason

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2waoh! that's really cool way back then, when i first learned this stuff, you used l'Hopital when you had no other way out. it was a footnote. so thanks :p

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0algebraic tricks aren't always easy to see so knowing l'hospital is a good back up plan or a first plan whatever
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