Lim x--> infinity (2x/x+1)^5

 

 

Can I get the limit into the brackets , apply l'hopital rule and the result ^5?

 

Then the answer is 2^5 ?

You can simplify by 2x/(x+1) = 2/(1 + 1/x), so you can get the limit directly.

Lim x--> infinity (2x/x+1)^5

 

 

Can I get the limit into the brackets , apply l'hopital rule and the result ^5?

 

Then the answer is 2^5 ?

To find the above limit you need the following theorem:

 

[math]lim_{x\to\infty} f(x)=m\Longrightarrow lim_{x\to\infty} [f(x)]^n = m^n[/math] for all natural Nos n

I think it's worth noting that all three suggestions so far do lead to the same correct result. The lesson to be learned perhaps, is that if L'Hopital needs to be applied five times then maybe it's worth looking for a quicker approach.