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anonymous
 4 years ago
\huge lim_{x rightarrow 1} frac{sqrt[5]{2x+1}1}{sqrt[4]{2x+1}1}
anonymous
 4 years ago
\huge lim_{x rightarrow 1} frac{sqrt[5]{2x+1}1}{sqrt[4]{2x+1}1}

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myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow 1}\frac{\sqrt[5]{2x+1}1}{\sqrt[4]{2x+1}1}\] Correct?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2Is so the function is continuous at x=1 so just plug it right on in :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@myininaya : sry i wrote it wrong ,Limit Tends to 0 .

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2Ok and hey which way do you prefer to do this? algebraically or l'hospital?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow 0}\frac{\sqrt[5]{2x+1}1}{\sqrt[4]{2x+1}1}\] Let me think on the algebraic approach I think it might be a substitution So the lcm of 5 and 4 is 20 So lets try the sub \[u^{20}=2x+1\] So that means \[(u^{20})^\frac{1}{5}=(2x+1)^\frac{1}{5} => u^4=\sqrt[5]{2x+1}\] and \[(u^{20})^\frac{1}{4}=(2x+1)^\frac{1}{4} => u^5=\sqrt[4]{2x+1}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2So if x goes to 0 then what does u go to ?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2\[u^{20}=2x+1 => u=(2x+1)^\frac{1}{20} \text{ correct?}\] So can you tell me what u goes to if x goes to 0?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2Eyad we are almost there :)
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