# Just want to verify my proof of a limit

• January 10th 2013, 03:49 PM
mathisfun26
Just want to verify my proof of a limit
$\lim_{x\rightarrow 2} \frac{x^2+x-6}{x-2}= 5$

Proof: given any $\epsilon >0$ choose $\delta =\epsilon$

Ignore the *******< br/ > *******

then if $\left | x-a \right | < \delta$
$\Rightarrow \left | x-2 \right |<\epsilon \Rightarrow \left | x+3-5 \right | <\epsilon
\Rightarrow \left | \frac{ (x+3)(x-2)}{x-2} -5 \right | <\epsilon \Rightarrow \left | \frac{x^2+x-6}{x-2}-5\right | < \epsilon \Rightarrow \left | f(x)-L \right | < \epsilon$

Therefore $\lim_{x\rightarrow 2} \frac{x^2+x-6}{x-2}= 5$
I want to make sure if I'm right thanks guys! This is just the proof no scratch work
• January 10th 2013, 04:02 PM
HallsofIvy
Re: Just want to verify my proof of a limit
Yes, that's a perfectly good proof. The basic idea, of course, is that, for x not equal to 2, $\frac{x^2+ x- 6}{x- 2}= \frac{(x+ 3)(x- 2)}{x- 2}= x+ 3$ so this problem is really "linear" and you can do it that way.
• January 10th 2013, 05:13 PM
Plato
Re: Just want to verify my proof of a limit
Quote:

Originally Posted by mathisfun26
$\lim_{x\rightarrow 2} \frac{x^2+x-6}{x-2}= 5$
Proof: given any $\epsilon >0$ choose $\delta =\epsilon$ then if $\left | x-a \right | < \delta$

Actually I would do this differently form reply #2.

Note that $\text{If }x\ne 2\text{ then }\left|\frac{x^2+x-6}{x-2}-5\right|=|x-2|$.

So you are correct we can use $\delta=\varepsilon.$
• January 10th 2013, 07:08 PM
Deveno
Re: Just want to verify my proof of a limit
Plato's reply points out a certain flaw in the original proof:

what if |x - 2| = 0?

if $0 < |x - 2| < \delta$....