Question

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

Answer 1

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

Answer 2

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

Answer 3

if not is *

Answer 4

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

Answer 5

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

Answer 6

algebraically

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Eyad we are almost there :)