## frx 3 years ago Let $f(x) = x^2 - 2x +1$ use the epsilon-delta definition to show that: $\lim_{x \rightarrow 3} f(x) = 4$ I've come up with this: $|( x-1)^2 - 4| < \epsilon$ $| x-3| < \sqrt{\epsilon }$ Set delta as the squareroot of epsilon: $| x-3| < \delta =\sqrt{\epsilon }$ How do I finish this, I get it all wrong by taking the left expresson to the second power

It can be done something like this $f(x)= x^2-2x+1$ and $\lim_{x\rightarrow3}x^2-2x+1=4$ first set $$\epsilon >0$$ Now find a $$\delta > 0$$ That relies on $$\epsilon$$ First we note that $$\left| (x^2-2x+1)-4\right| < \epsilon$$ can be stated like $$\left| x^2-2x-3 \right| < \epsilon$$ and then $$\left| x+1\right| \left| x-3\right| < \epsilon$$ Now we are actually looking for a delta, to do this we need to make use of the x+1 and get some sort of value for it so we can rewrite the above with the x-3 term. To do this we first make the statement that $$\delta \leq 1$$ this is always useful for these types of limit proofs because, A it gives us a number to work with, and B its a fair assumption as less than or equal to 1 is probably the GREATEST you would want in finding values for delta. Ok, so now lets get some values for x+1, given that $$| x-3| < \delta \leq 1$$ we can write $$|x-3| < 1$$ this implies $$-1 < x-3 < 1$$ which also implies $$2 < x <4$$ Ok, now we are in a great position, we have some values to use to change the x+1 to a number using the above. $$3 < | x+1| < 5$$ Now we can continute with our epsilon $$|x+1| |x-3| < |x-3| \cdot 5 < \epsilon$$ Giving us $$|x-3| < \frac{\epsilon}{5}$$ Which we can use for our delta ( providing that this value is < 1, if not then choose 1) $$\delta = \frac{\epsilon}{5}$$