Compute$$\lim_{x\to\infty}{3^x\over \sqrt {9^x - 4^x}}$$
Can one use end behavior to solve this? I.E.
$$\lim_{x\to\infty}{\sqrt{3^{2x}\over (3^{2x} - 2^{2x})}}$$
and therefore divide $3^{2x}$ by $3^{2x}$? leaving us with an anwser of $1$?
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2$\begingroup$ You can divide top and bottom by $3^{2x}$. Just make sure to divide all of the terms. $\endgroup$– J126Nov 23, 2014 at 17:30
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1$\begingroup$ That's right, you can. $\endgroup$– user65203Nov 23, 2014 at 17:31
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3 Answers
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HINT. Yes, note that: $${\sqrt{3^{2x}\over (3^{2x} - 2^{2x})}}={\sqrt{3^{2x}\over 3^{2x}\left(1 - \left(\frac{2}{3}\right)^{2x}\right)}}$$ but $\left(\frac{2}{3}\right)^{2x}\rightarrow 0$.
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For his problem exist simple solution.
$$ \lim_{x \to \infty^{+}} \frac{3^x}{\sqrt{9^x - 4^x}} = \lim_{x \to \infty^{+}} \frac{3^x}{\sqrt{(3^2)^x( 1 - \frac{4^x}{9^x})}} = \lim_{x \to \infty^{+}} \frac{3^x}{\sqrt{(3^x)^2} \cdot \sqrt{1 - \frac{4^x}{9^x}}} = \\ = \lim_{x \to \infty^{+}} \frac{3^x}{3^x \cdot \sqrt{1 - \left(\frac{4}{9}\right)^x}} = \lim_{x \to \infty^{+}} \frac{1}{\sqrt{1 - \left(\frac{4}{9}\right)^x}} = 1$$
I used here fact, that $|q| < 1 \Rightarrow \lim_{x \to \infty}q^x=0$. In fact it's same to yours, but I excluded $3^x$ from root, you just included.
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Hints:
$$\begin{align*}&(1)\;\;\sqrt{9^x-4^x}=3^x\sqrt{1-\left(\frac49\right)^x}\\{}\\&(2)\;\;|a|<1\implies a^x\xrightarrow[x\to\infty]{}0\end{align*}$$
