Just want to verify my proof of a limit

$\displaystyle \lim_{x\rightarrow 2} \frac{x^2+x-6}{x-2}= 5 $

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

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

then if $\displaystyle \left | x-a \right | < \delta $

$\displaystyle \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 $\displaystyle \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

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, $\displaystyle \frac{x^2+ x- 6}{x- 2}= \frac{(x+ 3)(x- 2)}{x- 2}= x+ 3$ so this problem is really "linear" and you can do it that way.

Re: Just want to verify my proof of a limit

Quote:

Originally Posted by

**mathisfun26** $\displaystyle \lim_{x\rightarrow 2} \frac{x^2+x-6}{x-2}= 5 $

Proof: given any $\displaystyle \epsilon >0 $ choose $\displaystyle \delta =\epsilon $ then if $\displaystyle \left | x-a \right | < \delta $

Actually I would do this differently form reply #2.

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

So you are correct we can use $\displaystyle \delta=\varepsilon.$

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?

your proof should start with:

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

this allows you to multiply and divide by (x - 2) without fear of it being undefined.