Limit w/ Sqrt function in numerator

• Sep 7th 2012, 07:43 PM
AZach
Limit w/ Sqrt function in numerator
$\displaystyle \lim_{x \to inf} \frac{\sqrt {2 + 9x^2}}{5+2x}$

I don't really know how to algebraically simplify the top. I tried to do the divide by x trick (since x is the variable with the greatest denominator available.

I'm not really sure how to get the LaTeX going for the multiple fraction bars yet, but it would look like the original problem with lim x->inf sqrt(2+9x^2/x^2) / 5+2x/x. I do the canceling and the top gives me sqrt(11) / 7, which my text doesn't agree with. I know about the conjugate radical trick, but that doesn't seem like an option here. I've attempted to square the top and bottom but I couldn't find a solution there. Is there a simple algebra trick that I can do to get ride of the square root on the top? All I can think know is that both the top and bottom are approaching infinity, but I can't deduce which is going faster (if either one is going father than the other).
• Sep 7th 2012, 08:00 PM
Prove It
Re: Limit w/ Sqrt function in numerator
Quote:

Originally Posted by AZach
$\displaystyle \lim_{x \to inf} \frac{\sqrt {2 + 9x^2}}{5+2x}$

I don't really know how to algebraically simplify the top. I tried to do the divide by x trick (since x is the variable with the greatest denominator available.

I'm not really sure how to get the LaTeX going for the multiple fraction bars yet, but it would look like the original problem with lim x->inf sqrt(2+9x^2/x^2) / 5+2x/x. I do the canceling and the top gives me sqrt(11) / 7, which my text doesn't agree with. I know about the conjugate radical trick, but that doesn't seem like an option here. I've attempted to square the top and bottom but I couldn't find a solution there. Is there a simple algebra trick that I can do to get ride of the square root on the top? All I can think know is that both the top and bottom are approaching infinity, but I can't deduce which is going faster (if either one is going father than the other).

Multiply the top and bottom by \displaystyle \displaystyle \begin{align*} \frac{1}{x} \end{align*}. This gives

\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty}\frac{\sqrt{2+9x^2}}{5+2x} &= \lim_{x \to \infty}\frac{\frac{1}{x}\sqrt{2+9x^2}}{\frac{1}{x} (5 + 2x)} \\ &= \lim_{x \to \infty}\frac{\sqrt{\frac{2 + 9x^2}{x^2}}}{\frac{5}{x} + 2} \\ &= \lim_{x \to \infty}\frac{\sqrt{\frac{2}{x^2} + 9}}{\frac{5}{x} + 2} \\ &= \frac{\sqrt{0 + 9}}{0 + 2} \\ &= \frac{3}{2} \end{align*}
• Sep 7th 2012, 08:17 PM
SworD
Re: Limit w/ Sqrt function in numerator
There's a nice intuitive way to quickly 'see' the limit value.

You probably learned that if two polynomials are of even degree, the limit of their quotient in infinity is the quotient of their leading coefficients. For example:

$\displaystyle \lim_{x \to \infty} \frac{2x^2 + 5x + 2}{5+3x^2} = \frac{2}{3}$

You can apply the same concept here. As x --> infinity, the 2 in the numerator term matters less and less, swamped by the 9x^2 term. Therefore your limit can be simplified to:

$\displaystyle \lim_{x \to \infty} \frac{\sqrt {2 + 9x^2}}{5+2x} = \lim_{x \to \infty} \frac{\sqrt {9x^2}}{5+2x} = \lim_{x \to \infty} \frac{3x}{2x} = \frac{3}{2}$
• Sep 7th 2012, 08:23 PM
AZach
Re: Limit w/ Sqrt function in numerator
Quote:

Originally Posted by Prove It
\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty}\frac{\sqrt{2+9x^2}}{5+2x} &= \lim_{x \to \infty}\frac{\frac{1}{x}\sqrt{2+9x^2}}{\frac{1}{x} (5 + 2x)} \\ &= \lim_{x \to \infty}\frac{\sqrt{\frac{2 + 9x^2}{x^2}}}{\frac{5}{x} + 2} \\ &= \lim_{x \to \infty}\frac{\sqrt{\frac{2}{x^2} + 9}}{\frac{5}{x} + 2} \\ &= \frac{\sqrt{0 + 9}}{0 + 2} \\ &= \frac{3}{2} \end{align*}

Thanks for both the responses. I realized in the second step listed I just crossed out the x^2 altogether, is that wrong mathematically? Sorry for quoting the entire derivation, but I didn't want to trim it down because I didn't want to break the code...