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I'm new to $\delta, \epsilon$ proofs and not sure if I've got the hand of them quite yet.

$$ \lim_{x\to -2} (2x^2+5x+3)= 1 $$

$|2x^2 + 5x + 3 - 1| < \epsilon$

$|(2x + 1)(x + 2)| < \epsilon$

$|(2x + 4 - 3)(x + 2)| < \epsilon$

$|(2(x+2)^2 -3(x + 2)| \leq 2|x+2|^2 +3|x + 2| < \epsilon$ (via the triangle inequality)

Let $|x+2| < 1$

Then $\delta=\min\left(\dfrac{\epsilon}{5}, 1\right)$

and

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

$|3x^2 + 4x - 2 - 2| < \epsilon$

$|3x^2 - 12 + 4x + 8| < \epsilon$

$|3(x+2)(x-2) + 4(x + 2)| < \epsilon$

$|3(x+2)(x + 2 -4) + 4(x + 2)| < \epsilon$

$|3[(x+2)^2 -4(x+2)] + 4(x + 2)| < \epsilon$

$|3(x+2)^2 - 8(x+2)| \leq 3|x+2|^2 + 8|x+2| < \epsilon$

Let $|x + 2| < 1$

Then $\delta =\min\left(\dfrac{\epsilon}{11}, 1\right)$

and to make sure I'm understanding this properly, when we assert that $|x+2| < 1$, this means that $\delta \leq 1$ as well, because if $\delta \geq 1$, this would allow for $|x+2| \geq 1$, which violates that condition we just imposed?

Edit: Apologies for the bad tex

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    See [here](http://math.stackexchange.com/questions/209440/how-to-show-that-fx-x2-is-continuous-at-x-1/209492#209492).2012-10-26

1 Answers 1