# Limit problem

• February 3rd 2011, 02:56 PM
polarbear73
Limit problem
I hope this is the right forum for this. The book says that the limit of $(x)\frac{4x}{(x-4)^2{x}}$ as $x\rightarrow 0$ is $\frac{1}{4}$ I've stared at it and just can't see how I avoid having an x term in the numerator, and thus the limit being zero. The problem is, this book has been shown to be wrong at least once before, so now I'm never sure. If the limit is $\frac{1}{4}$ can someone show me why?
• February 3rd 2011, 03:07 PM
pickslides
$\displaystyle \lim_{x\to 0}\frac{4x}{(x-4)^2{x}}$

Cancelling the $x$ 's

$\displaystyle \lim_{x\to 0}\frac{4}{(x-4)^2}$

$\displaystyle \frac{4}{((0)-4)^2}$

$\displaystyle \frac{4}{16} = \frac{1}{4}$
• February 3rd 2011, 05:40 PM
polarbear73
Sorry, I don't think I presented the problem correctly. I don't know how to do that limit notation with the arrow under it. The problem arose from considering the degree of singular points. I was using the test of the limit of the quantity $(x-x_{0})\frac{4x}{(x-4)^2{x}}$ where $x_0=0$. The fraction is the p(x) value of a second order ODE. Is the limit still 1/4?
• February 3rd 2011, 06:06 PM
Prove It
Quote:

Originally Posted by polarbear73
I don't know how to do that limit notation with the arrow under it.

\lim_{x \to h}f(x) gives $\displaystyle \lim_{x \to h}f(x)$
• February 4th 2011, 02:33 AM
polarbear73
Thanks for that information.