# Prove this limit

• Jun 9th 2012, 03:57 PM
sluggerbroth
Prove this limit
lim as x goes to 1 (x^(2)-2x+4)=3
• Jun 10th 2012, 11:56 AM
emakarov
Re: Prove this limit
If a function f is continuous at x0, then its limit as x -> x0 is f(x0).
• Jun 10th 2012, 03:23 PM
HallsofIvy
Re: Prove this limit
Using what "basis"? If you know about "continuous functions" and in particular that all polynomials are continous for all values of x then it is sufficient to say, as emakarov does, that the limit is $\displaystyle (1)^2-2(1)+ 4= 1- 2+ 4= 3$.

If, on the other hand, you have only the definition of limit to work with, you need to look at $\displaystyle |f(x)- L|= |x^2- 2x+ 4- 3|= |x^2- 2x+ 1|= |x-1|^2< \epsilon$, then what can you say if you choose $\displaystyle \delta= \sqrt{\epsilon}$ so that $\displaystyle |x- 1|< \delta$ becomes $\displaystyle |x- 1|< \sqrt{\epsilon}$?
• Jun 14th 2012, 11:54 AM
richard1234
Re: Prove this limit
The function $\displaystyle f(x) = x^2 - 2x + 4$ is continuous everywhere.

Therefore $\displaystyle \lim_{x \to 1} (x^2 - 2x + 4) = (1)^2 - 2(1) + 4 = 3$