ahmed sobhy Posted January 7, 2012 Posted January 7, 2012 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 ?
mathematic Posted January 7, 2012 Posted January 7, 2012 You can simplify by 2x/(x+1) = 2/(1 + 1/x), so you can get the limit directly.
kavlas Posted January 9, 2012 Posted January 9, 2012 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
the tree Posted January 9, 2012 Posted January 9, 2012 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.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now