# How to take limit of a function from one side

• Nov 6th 2012, 05:44 PM
PhizKid
How to take limit of a function from one side
$\lim_{x \to 1^{+} } \frac{-x}{x - 2}$

I only know how to do this by direct substitution, but I'm not sure how to do it from one side. In the calculator, I tried putting in:

-1.000001 / (1.000001 - 2) and it gives me a positive value of a little greater than 1, but I don't know how to algebraically show this step since saying "I put a value close to 1 a little greater than 1 into the calculator" isn't a valid response, I think.
• Nov 6th 2012, 06:19 PM
HallsofIvy
Re: How to take limit of a function from one side
I'm not sure I understand your question. $\frac{-x}{x- 2}$ is a rational function so that it is "continuous" where it is defined. In particular, setting x= 1, $\frac{-1}{1- 2}= \frac{-1}{-1}= 1$. Since that exists, the function is continuous and its limit, as x goes to 1, is the value of the function, 1. And since the limit exists, the two one sided limits must be the same, 1.

If you don't like that, you could copy the proof of the limit, restricting x to be larger than 2: Given $\epsilon> 0$, we need to find $\delta> 0$, such that if $x- 2< \delta$ (we don't need the absolute value because we are assuming x> 2) then $\left|\frac{-x}{x- 2}- 1\right|< \epsilon$. $\frac{-x}{x- 2}- 1= \frac{-x-x+ 2}{x- 2}= \frac{-2x+ 2}{x- 2}$ so we have $\left|\frac{-x}{x-2}- 1\right|= 2\left|\frac{x- 1}{x- 2}$. If x is larger than 1 but close to it, we can assume 1/2< x< 3/2 so that 1/2- 2= -3/2< x- 2< 3/2- 2= -1/2 and |x- 2|> 1/2. With that, if $x- 1< \delta= \epsilon$ then $\left|\frac{-x}{x-2}-1\right|< \epsilon$.