Prove this limit

**sluggerbroth** · June 9th 2012, 03:57 PM

lim as x goes to 1 (x^(2)-2x+4)=3

**emakarov** · June 10th 2012, 11:56 AM

If a function f is continuous at x0, then its limit as x -> x0 is f(x0).

**HallsofIvy** · June 10th 2012, 03:23 PM

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 $(1)^2-2(1)+ 4= 1- 2+ 4= 3$ .

If, on the other hand, you have only the definition of limit to work with, you need to look at $|f(x)- L|= |x^2- 2x+ 4- 3|= |x^2- 2x+ 1|= |x-1|^2< \epsilon$ , then what can you say if you choose $\delta= \sqrt{\epsilon}$ so that $|x- 1|< \delta$ becomes $|x- 1|< \sqrt{\epsilon}$ ?

**richard1234** · June 14th 2012, 11:54 AM

The function $f(x) = x^2 - 2x + 4$ is continuous everywhere.

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

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