Limit Question: sqrt[9 + 3x^2]/(2 + 7x), x -> infinity

[killasnake]

With this limit question

I take the highest power on each line and divide that right? so...

3/7 ?

 

[soroban]

Re: Limit Question

Hello, killasnake!

\(\displaystyle \L \lim_{x\to\infty}\frac{\sqrt{9\,+\,3x^2}}{2\,+\,7x}\)

Divide top and bottom by $\displaystyle x$.

. . \(\displaystyle \L\frac{\frac{\sqrt{9\,+\,3x^2}}{x}} {\frac{2\,+\,7x}{x}} \;=\;\frac{\frac{\sqrt{9\,+\,3x^2}}{\sqrt{x^2}}}{\frac{2\,+\,7x}{x}} \;=\;\frac{\sqrt{\frac{9\,+\,3x^2}{x^2}}}{\frac{2\,+\,7x}{x}} \;=\;\frac{\sqrt{\frac{9}{x^2}\,+\,3}}{\frac{2}{x}\,+\,7}\)

Therefore:\(\displaystyle \L\:\lim_{x\to\infty}\frac{\sqrt{\frac{9}{x^2}\,+\,3}}{\frac{2}{x}\,+\,7} \;=\; \frac{\sqrt{0\,+\,3}}{0\,+\,7}\;=\;\frac{\sqrt{3}}{7}\)

 

[galactus]

You're close.

\(\displaystyle \L\\\lim_{x\to\infty}\frac{\sqrt{3(3+x^{2})}}{2+7x}\)

\(\displaystyle \L\\\sqrt{3}\lim_{x\to\infty}\frac{\sqrt{3+x^{2}}}{2+7x}\)

Divide numerator by $\displaystyle \sqrt{x^{2}}$ and denominator by $\displaystyle x$

\(\displaystyle \L\\\sqrt{3}\lim_{x\to\infty}\frac{\sqrt{\frac{3}{x^{2}}+1}}{\frac{2}{x}+7}\)

Now, see what the limit is as x approaches infinity?.

\(\displaystyle \L\\\sqrt{3}\lim_{x\to\infty}\frac{\sqrt{1}}{7}\)

\(\displaystyle =\H\\\frac{\sqrt{3}}{7}\)


EDIT: You ornery Soroban, you beat me. Oh well, my approach is slightly different.

 

[killasnake]

ohh... gotcha Thank you for the help, soroban and galactus