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

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

EyadBest ResponseYou've already chosen the best response.0
@myininaya : sry i wrote it wrong ,Limit Tends to 0 .
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
Ok and hey which way do you prefer to do this? algebraically or l'hospital?
 one year ago

myininayaBest 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 \[(u^{20})^\frac{1}{4}=(2x+1)^\frac{1}{4} => u^5=\sqrt[4]{2x+1}\]
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
So if x goes to 0 then what does u go to ?
 one year ago

myininayaBest 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?
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
Eyad we are almost there :)
 one year ago
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