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

1. myininaya

$\lim_{x \rightarrow 1}\frac{\sqrt[5]{2x+1}-1}{\sqrt[4]{2x+1}-1}$ Correct?

2. myininaya

Is so the function is continuous at x=1 so just plug it right on in :)

3. myininaya

if not is *

4. anonymous

@myininaya : sry i wrote it wrong ,Limit Tends to 0 .

5. myininaya

Ok and hey which way do you prefer to do this? algebraically or l'hospital?

6. anonymous

algebraically

7. myininaya

$\lim_{x \rightarrow 0}\frac{\sqrt[5]{2x+1}-1}{\sqrt[4]{2x+1}-1}$ 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 $(u^{20})^\frac{1}{4}=(2x+1)^\frac{1}{4} => u^5=\sqrt[4]{2x+1}$

8. myininaya

So if x goes to 0 then what does u go to ?

9. myininaya

$u^{20}=2x+1 => u=(2x+1)^\frac{1}{20} \text{ correct?}$ So can you tell me what u goes to if x goes to 0?

10. myininaya

Eyad we are almost there :)