6
$\begingroup$

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 $1$?

$\endgroup$
2
  • 2
    $\begingroup$ You can divide top and bottom by $3^{2x}$. Just make sure to divide all of the terms. $\endgroup$
    – J126
    Nov 23, 2014 at 17:30
  • 1
    $\begingroup$ That's right, you can. $\endgroup$
    – user65203
    Nov 23, 2014 at 17:31

3 Answers 3

5
$\begingroup$

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 $\left(\frac{2}{3}\right)^{2x}\rightarrow 0$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Hints:

$$\begin{align*}&(1)\;\;\sqrt{9^x-4^x}=3^x\sqrt{1-\left(\frac49\right)^x}\\{}\\&(2)\;\;|a|<1\implies a^x\xrightarrow[x\to\infty]{}0\end{align*}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.