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frx
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Let \[f(x) = x^2  2x +1 \] use the epsilondelta definition to show that: \[\lim_{x \rightarrow 3} f(x) = 4\]
I've come up with this:
\[( x1)^2  4 < \epsilon \]
\[ x3 < \sqrt{\epsilon } \]
Set delta as the squareroot of epsilon:
\[ x3 < \delta =\sqrt{\epsilon } \]
How do I finish this, I get it all wrong by taking the left expresson to the second power
 one year ago
 one year ago
frx Group Title
Let \[f(x) = x^2  2x +1 \] use the epsilondelta definition to show that: \[\lim_{x \rightarrow 3} f(x) = 4\] I've come up with this: \[( x1)^2  4 < \epsilon \] \[ x3 < \sqrt{\epsilon } \] Set delta as the squareroot of epsilon: \[ x3 < \delta =\sqrt{\epsilon } \] How do I finish this, I get it all wrong by taking the left expresson to the second power
 one year ago
 one year ago

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genix Group TitleBest ResponseYou've already chosen the best response.0
It can be done something like this \[f(x)= x^22x+1 \] and \[\lim_{x\rightarrow3}x^22x+1=4 \] first set $$\epsilon >0$$ Now find a $$\delta > 0$$ That relies on $$\epsilon$$ First we note that $$ \left (x^22x+1)4\right < \epsilon$$ can be stated like $$ \left x^22x3 \right < \epsilon $$ and then $$ \left x+1\right \left x3\right < \epsilon $$ 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 x3 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 $$  x3 < \delta \leq 1$$ we can write $$ x3 < 1 $$ this implies $$ 1 < x3 < 1 $$ 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 $$ x+1 x3 < x3 \cdot 5 < \epsilon$$ Giving us $$ x3 < \frac{\epsilon}{5} $$ Which we can use for our delta ( providing that this value is < 1, if not then choose 1) $$ \delta = \frac{\epsilon}{5}$$
 one year ago
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