# Limits (L'Hopital's Rule)

#### n3rdwannab3

**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
$$\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}\)

#### n3rdwannab3

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
Inside the brackets, multiply by (1/9x)/(1/9x) (which equals 1).

#### n3rdwannab3

Okay I understand now, thank you very much!