2
$\begingroup$

Would anyone know how to prove the limit of this cubic equation using the epsilon delta definition?

$\lim_{x \rightarrow 2} x^3 +2x^2 -x -1 = 13 $ I really don't know where to start other than inputting the values of $a$, $L$ and $f(x)$ for this example into the definition of a limit:

$0 < |x - 2| < d $ implies $|(x^3 +2x^2 -x -1) - 13| < \epsilon$

$\endgroup$
  • $\begingroup$ $x^3+2x^2-x-1-13=(x-2)Q(x)\\=(x-2)(x^2+4x+7)\\$ so put at $|(x^3 +2x^2 -x -1) - 13| < \epsilon\\|(x-2)||x^2+4x+7|< \epsilon$ $\endgroup$ – Khosrotash Oct 16 '16 at 10:51
2
$\begingroup$

Let $f(x)=x^3+2x^2-x+1. $ Note $|f(x)-13|=|x-2||x^2+4x+7|$

For $0<|x-2|<1$, we have $1<x<3\Rightarrow12<x^2+4x+7<28$,

Let $\epsilon>0$, then take $\delta(\epsilon):=\min\{\epsilon/28,1\}$. For $0<|x-2|<\delta(\epsilon), $ we have

$$|f(x)-13|=|x-2||x^2+4x+7|<\frac{\epsilon}{28}28=\epsilon$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.