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Use the definition of a limit to prove the following:

$$\lim_{x\to -2}(x-3x^2)=-14.$$

Our definition: Let $L$ be a number and let $f(x)$ be a function which is defined on an open interval containing $c$, expect possibly not at $c$ itself. If for every $\varepsilon>0$ there exists a corresponding $\delta>0$ such that $o<|x-c|<\delta \Rightarrow |f(x)-L|<\varepsilon$ then we say $f(x)$ has a limit $L$ as $x$ approaches $c$.

Not really sure how I go about doing this.

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  • $\begingroup$ You should thoroughly study the linked examples on the sidebar to the right (where it says "Related"). For example, this one $\endgroup$ – The Chaz 2.0 Oct 21 '13 at 2:45
  • $\begingroup$ Hint: replace the $c$ in your definition with the value $x$ is approaching in your problem. $\endgroup$ – Jamil_V Oct 21 '13 at 2:46
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We want to show that for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that $|(x-3x^2)-(-14)|\lt \epsilon$ whenever $|x-(-2)|\lt \delta$. So suppose that we are given an $\epsilon$. We show how to produce a suitable $\delta$. Note that $$(x-3x^2)-(-14)=(x+2)-3(x^2-4)=(x+2)(7-3x).$$ Thus $$|(x-3x^2)-(-14)|=|x+2||7-3x|.$$ First of all, we will make sure that $-3\lt x\lt -1$, by picking $\delta\lt 1$. Then $|7-3x|\lt 16$. Also make sure that $\delta\lt \frac{\epsilon}{16}$. So to sum up, we choose $\delta=\min(1,\frac{\epsilon}{16})$. If $|x+2|\lt \delta$, then $|(x-3x^2)-(-14)|\lt \epsilon$.

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  • $\begingroup$ I dont really understand the algebra behind that, why is (x+2)−3(x2−4)=(x+2)(7−3x)? $\endgroup$ – Arnold Oct 21 '13 at 4:46
  • $\begingroup$ We have $x^2-4=(x+2)(x-2)$. So taking the common factor $x+2$ "out" we get $(x+2)(1-3(x-2))$. $\endgroup$ – André Nicolas Oct 21 '13 at 4:56

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