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Could you help me show that

$$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$$

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The short, conceptual answer is that the terms that dominate the growth of the numerator and denominator are both $3^{2n}$. – orlandpm Dec 29 '12 at 22:29
up vote 8 down vote accepted

Divide numerator and denominator by $3^{2n}$, yielding:


Prove the numerator and denominator both approach $1$ as $n\to\infty$.

This amounts to showing $\frac{n^2}{3^{2n}}\to 0$ and $\frac{n^3}{3^n}\to 0$ as $n\to\infty$.

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The standard approach works: multiply numerator and denominator by the reciprocal of the fastest growing term in the denominator to get

$$\begin{align*} \frac{n^2+3^{2n}}{\left(n^3+3^n\right)^2}&=\frac{n^2+3^{2n}}{n^6+2n^33^n+3^{2n}}\\ &=\frac{n^2+3^{2n}}{n^6+2n^33^n+3^{2n}}\cdot\frac{1/3^{2n}}{1/3^{2n}}\\ &=\frac{\dfrac{n^2}{3^{2n}}+1}{\dfrac{n^6}{3^{2n}}+\dfrac{2n^3}{3^n}+1}\;. \end{align*}$$

Now take the limit as $n\to\infty$.

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I did a similar approach. Unfortunately I took the wrong therm $n^3$ instead $3^{2n}$. Is there a good resource that explains the 'growth' of functions? – leo Dec 29 '12 at 22:19
Most discrete math texts have some material on the subject. However, you’ll be able to get by pretty well if you know that $x^b$ grows faster than $x^a$ when $b>a>0$, that $a^x$ grows faster than than any $x^b$ if $a>1$, and that $\log_a x$ grows slower than any $x^b$ if $a>1$. – Brian M. Scott Dec 29 '12 at 22:25

$\displaystyle \frac{n^2 + 3^{2n}}{(n^3 + 3^n)^2} = \frac{n^2}{(n^3 + 3^n)^2} + \frac{3^{2n}}{(n^3 + 3^n)^2}$.

Since $\frac{n^2}{(n^3 + 3^n)^2} \to 0$ as $n\to \infty$, it's enough to show that $\displaystyle \lim_{n \to \infty} \frac{3^{2n}}{(n^3 + 3^n)^2} = 1$.

To see this, write $ \displaystyle \frac{3^{2n}}{(n^3 + 3^n)^2} = \frac{3^{2n}}{(3^n(\frac{n^2}{3^n} + 1))^2} = \frac{1}{(\frac{n^2}{3^n} + 1)^2}$.

Observing that $\frac{n^2}{3^n} \to 0$ as $n \to \infty$ and the usual statements of continuity, you're done.

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$$\dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = \frac{1 + \frac{n^2}{3^{2n}}}{\left( 1 + \frac{n^3}{3^{2n}}\right)^2}$$ As exponential grows more faster than polynomial, you have $1/1$

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$$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} =\lim_{n\to\infty}{n^2+3^{2n}\over n^6+2n^33^n+3^{2n}}=\lim_{n\to\infty}{\frac{n^2}{3^{2n}}+1\over\frac{n^6}{3^{2n}}+2\frac{n^3}{3^n}+1}=1$$

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