Limits (L'Hopital's Rule)

Nov 2009
37
0
Please see attached image

**edit** I recognize this as an indeterminate power problem but I don't know how to apply the hint to solve the problem
 

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greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
\(\displaystyle \lim_{x\to\infty}\,\(\frac{9x}{9x\,+\,6}\)^{7x}\,=\,\lim_{x\to\infty}\,\frac{1}{\(1\,+\,\frac{6}{9x}\)^{7x}}\,=\,\lim_{x\to\infty}\,\frac{1}{\[\(1\,+\,\frac{2}{3x}\)^{x}\]^7}\,=\,\frac{1}{\(e^{2/3}\)^7}\,=\,e^{-14/3}\)
 
Nov 2009
37
0
Hi Greg, thanks very much for your help, but how did you get to the first step? (below)

\(\displaystyle \lim_{x\to\infty}\,\(\frac{9x}{9x\,+\,6}\)^{7x}\,=\,\lim_{x\to\infty}\,\frac{1}{\(1\,+\,\frac{6}{9x}\)^{7x}}\)

Sorry but I just started learning calculus and it is very difficult for me
 

greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
Inside the brackets, multiply by (1/9x)/(1/9x) (which equals 1).