calculate the limit of the following functions and prove directly from definition (using $\epsilon$ and $\delta$)
a) $$\lim _{x\to 2}(x^3+3x)$$

We just learned the definition of limits in regards to function and this is supposed to be a simple question. But I just can't understand how to use the info I'm given.

a) Am I right with guessing the limit is $2^3+3\cdot 2=14$?

b) Assuming I am (I don't think it would change much if I am not), then for an $\epsilon$ I want to find a $\delta$ such that for each $x$ in the punctured environment of $(x_0 - \delta,x_0+\delta)$ , $|x^3+3x-14|<\epsilon$, but I have no idea how to proceed from here. I tried searching for similar examples to see how the truth process is done but wasn't able to find any.

  • $\begingroup$ Since the function $f(x)=x^3+3x$ is continuous at $2$ then the limit is $14$. $\endgroup$ – Sigur Dec 23 '12 at 13:32

a) You are right because $f(x)=x^3+3x$ is continuous and so $\lim_{x\to 2}f(x)=f(2)$

b) Observe that $$\left|x^3+3x-2^3-3\cdot 2\right|=\left|(x-2)(x^2+2x+4)+3(x-2)\right|= \left|x-2\right|\left|x^2+2x+7\right|<\delta\left|x^2+2x+7\right| $$ Can you proceed now?

EDIT: The term $\left|x^2+2x+7\right|$ is problematic and needs to be replaced by something containing only $\delta$. With $\left|x-2\right|<\delta$ in mind, $$\left|x^2+2x+7\right|=\left|(x-2)^2+6x+3\right|\le \left|x-2\right|^2+6\left|x+\frac12\right|<\delta^2+6\left|x-2+\frac52\right|<\\ \delta^2+6\delta+15$$ We have $$\delta\left|x^2+2x+7\right|<\delta^3+6\delta^2+15\delta$$ and want this to be less than $\epsilon$. If we further demand that $\delta<1$, $$\delta\left|x^2+2x+7\right|<\delta^3+6\delta^2+15\delta<\delta+6\delta+15\delta=18\delta$$ Taking $\delta<\frac{\epsilon}{18}$ yields the result (in fact we can take $$\delta=\frac12\min\left\{1,\frac{\epsilon}{18}\right\}$$ as our $\delta$ to be completely rigorous)

  • $\begingroup$ Both solutions really helped me. From what you gave me I was able to realize what my delta needs to be, and the solution below allowed me to understand how I'm supposed to write a proof on these matter. Thank you both! $\endgroup$ – Nescio Dec 23 '12 at 13:54
  • $\begingroup$ @Nescio I updated my answer with my take on how one can complte the proof $\endgroup$ – Nameless Dec 23 '12 at 14:02

You can evaluate the limit by substituting $2$ because $x^3+3x$ is continuous. To show that the limit is $14$, Let $\epsilon>0$. Let $\delta=\min\{1,\epsilon(\frac{1}{22})$. Now for all $x$ such that $|x-2|<\delta$, we have: $$|x^3+3x-14|\leq|x^3-8|+|3x-6|\leq|x^2+2x+4||x-2|+3|x-2|$$ $$|x^3+3x-14|<\delta (|x^2+2x+4|+3)$$ Since for all $x\in [2-\delta,2+\delta]\subseteq [1,3]$ we know that $|x^2+2x+4|+3\leq22$ Thus: $$|x^3+3x-14|<\delta (|x^2+2x+4|+3)<\min\{1,\epsilon(\frac{1}{22})\}(22)<\epsilon$$


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