• anonymous
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
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert
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  • katieb
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  • anonymous
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) $$ \delta = \frac{\epsilon}{5}$$

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