$\displaystyle \lim_{x\rightarrow\infty}f(x) = e^{\lim_{x\rightarrow\infty}ln(f(x))}$

I think I've seen or used this before but it might've just been something similar, I don't have my calculus book handy.

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- Apr 22nd 2010, 05:47 PMHD09Is this true?
$\displaystyle \lim_{x\rightarrow\infty}f(x) = e^{\lim_{x\rightarrow\infty}ln(f(x))}$

I think I've seen or used this before but it might've just been something similar, I don't have my calculus book handy. - Apr 22nd 2010, 05:52 PMDrexel28
If the limit exists, yes. The quick way using only continuity is that $\displaystyle \lim_{x\to\infty}f(x)=\lim_{z\to 0}f\left(\frac{1}{z}\right)=\lim_{z\to 0}\text{exp}\left(\ln\left(f\left(\frac{1}{z}\righ t)\right)\right)=\text{exp}\left(\lim_{z\to 0}\ln\left(f\left(\frac{1}{z}\right)\right)\right)$ and work backwards.