# limits pass papers

• Mar 18th 2012, 04:08 PM
mathkid12
limits pass papers
$\lim_{x \to +\4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .$
• Mar 18th 2012, 04:13 PM
Xeritas
Re: limits pass papers
So...
$\displaystyle \ \lim_{x \to \4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .\]$
You forgot the TEX tags.

The overall limit is indeterminate, although from the right side, denoted by $\displaystyle \lim_{x \to 4+}$, it should be infinity.
(Note the plus is on the other side of the four)

The easiest way to do this is to graph it then approximate the value by plugging in an obscure number such as 4.0000000000000001 instead of 4, this should give you some idea.
• Mar 18th 2012, 10:25 PM
princeps
Re: limits pass papers
Quote:

Originally Posted by mathkid12
$\lim_{x \to +\4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .$

$\displaystyle \displaystyle \lim_{x \to 4^{+}} \frac{\sqrt[2]{x}}{x^2 - 5x + 4}=\displaystyle \lim_{x \to 4^{+}} \frac {\sqrt[2]{x}}{(x-4)(x-1)}=\frac{2}{0^{+} \cdot 3} = \frac{2}{0^{+}}=+\infty$