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How do I find the limit of something like

$$ \lim_{x\to \infty} \frac{2\cdot3^{5x}+5}{3^{5x}+2^{5x}} $$

?

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Thanks everyone I think I understand how both infty's go to 0 now, remembering 1/little = big. –  eee3 Oct 27 '12 at 18:39

4 Answers 4

up vote 2 down vote accepted

Hint: $2\cdot 3^{5x}+5=3^{5x}\cdot (2+\frac5{3^{5x}})$ and $3^{5x}+ 2^{5x}=3^{5x}\cdot(1+(\frac23)^{5x})$.

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Why the downvote? –  Hagen von Eitzen Oct 20 '12 at 9:04

Divide both the upper and the lower term by $3^{5x}$, that will do it.

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Like Hagen von Eitzen's this answer was downvoted, why? –  filmor Oct 20 '12 at 10:39

Note that

$$\frac{{2 \cdot {3^{5x}} + 5}}{{{3^{5x}} + {2^{5x}}}} \sim \frac{{2 \cdot {3^{5x}}}}{{{3^{5x}} + {2^{5x}}}} = \frac{2}{{1 + {{\left( {\frac{2}{3}} \right)}^{5x}}}}.$$

So ...

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More generally, if you're trying to determine limiting behavior of a function of form $$\frac{f(x)}{g(x)}$$ as $x\to\infty$, and the limit is of form "$\pm\frac\infty\infty$", then you can look for a dominating term--that is, a term that (eventually) grows more rapidly in size than any other as $x$ gets sufficiently large--and then divide top and bottom by that dominating term (as filmor and Hagen demonstrated in their answers), so that all but a few terms vanish in the limit.

Some guidelines:

If $f$ and $g$ are linear combinations of nonnegative powers of $x$, then the dominating term is the highest power of $x$ that appears in the expression.

If $f$ and $g$ are linear combinations of exponential expressions, then the dominating term will be the one with the largest base (and in case of a tie, the one with the fastest growing power).

Exponential terms (with bases greater than $1$) dominate powers of $x$, and both of these dominate logarithmic terms.

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