# Math Help - limits pass papers

1. ## limits pass papers

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

2. ## Re: limits pass papers

So...
$\ \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 $\lim_{x \to 4+}$, 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.

3. ## Re: limits pass papers

Originally Posted by mathkid12
$\lim_{x \to +\4} = \frac{\sqrt[2]{x}}{x^2 - 5x + 4} .$
$\displaystyle \lim_{x \to 4^{+}} \frac{\sqrt[2]{x}}{x^2 - 5x + 4}=\displaystyle \lim_{x \to 4^{+}} \frac {\sqrt[2]{x}}{(x-4)(x-1)}=\frac{2}{0^{+} \cdot 3} = \frac{2}{0^{+}}=+\infty$