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i am pulling my hair out in solving this problem. i know, it is a stupid question but i am not that good at maths, and many thanks for any help

$\lim_{n \to \infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n}$

my steps are: $\lim_{n \to \infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n} = \frac{2^{n}2+3^{n}3}{2^n+3^n} = help = ? $ i know, it diverges as the denominator grows faster than the nominator, but the calculation is killing me now

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    Hint: divide each term by $3^{n+1}$.2012-12-15

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The main idea here and in similar problems is to factor out the terms not with the biggest exponents but with the biggest bases. That way you will create terms of the form $(a/b)^n$ where $a that converge to $0$. $\lim_{n \to \infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n}=\lim_{n \to \infty} \frac{3^{n+1}((\frac23)^{n+1}+1)}{3^n((\frac23)^n+1)}=\lim_{n \to \infty} 3\frac{(\frac23)^{n+1}+1}{(\frac23)^n+1}=3 $ because $(\frac{2}{3})^n\to 0$

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    @BabakSorouh yes, exactly. thanks2012-12-15
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Your first step is good. Now divide top and bottom by $3^n$.

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    The same strategy is useful for $\frac{2n^5+4n^3-1000n}{3n^5-99n^4-n^2+17}$, and many others, such as replacing top and bottom by their square roots.2012-12-15