ahmed sobhy Posted January 7, 2012 Share 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 ? Link to comment Share on other sites More sharing options...
mathematic Posted January 7, 2012 Share Posted January 7, 2012 You can simplify by 2x/(x+1) = 2/(1 + 1/x), so you can get the limit directly. Link to comment Share on other sites More sharing options...
kavlas Posted January 9, 2012 Share 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 Link to comment Share on other sites More sharing options...
the tree Posted January 9, 2012 Share 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. Link to comment Share on other sites More sharing options...
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