Question about continuity and the Epsilon-Delta definition So, I' currently working through an example:
Show that $\lim \limits_{x \to 0}\frac{x^3}{x-4}\cos{(\frac{1}{x})}=0$ using the $\epsilon-\delta$ definition.
So far I have the following working:
For any $\epsilon>0$, we must find a $\delta>0$ such that if $|x-0|<\delta$ then:
$|\frac{x^3}{x-4}\cos{(\frac{1}{x})}-0|<\epsilon, \qquad \forall x \in(-\delta,0)\cup(0,\delta)$
I know that the cosine function is bound by $[-1,1]$, so we can say:
$|\frac{x^3}{x-4}\cos{(\frac{1}{x})}-0|=|\frac{x^3}{x-4}||cos{(\frac{1}{x})}|\leq|\frac{x^3}{x-4}|<\epsilon$
Now the trouble I'm having is developing an upper bound for $|\frac{x^3}{x-4}|$. My suspicion is that since we a looking for a small neighbour hood of $\delta$; if I bind $\delta\leq1$ then $|x^3|$ should also be bounded the same way implying: $-1<x<1$.
My gut is telling me that $\delta=\epsilon$ is going to be the road I need to go down, but I want to prove it. Are there any insights I'm missing? Should I be placing an lower bound on $\frac{1}{x-4}$ by putting a positive upper bound on $x-4$?
Thanks for reading.
 A: Since $|x-4|\geq 4 -|x|$ we get $|\frac {x^{3}} {x-4}|\leq \frac {|x|^{3}} {4-|x|} <\epsilon$ if $|x|^{3}+\epsilon|x| <4\epsilon$. If $|x|<1$ we can use $|x|^{3} <|x|$ to reduce this to $|x| <4\epsilon/(1+\epsilon)$. We can take $\delta=\min \{1, 4\epsilon/(1+\epsilon)\}$.
A: Let's finish it based on your ideas.
As you said, choosing $\delta\leq1$ is a good idea. Notice that it gives us that
$$\lvert x^2\rvert< 1,$$
$$\lvert x-4\rvert\geq\lvert \lvert x\rvert-4\rvert=4-\lvert x\rvert>3.$$
Using this we have that
$$\left\lvert\frac{x^3}{x-4}\right\rvert<\frac{\lvert x\rvert }{3}<\lvert x\rvert.$$
Clearly, as you guessed yourself, $\delta=\varepsilon$ works wonders here. Thus, with
$$\delta=\min\left\{1,\varepsilon\right\}$$
we are done.
A: The first answer is correct. But this one goes (i believe), more in your original direction, but without restricting $\delta$.
In this first part $x<4$ :
For $f = |\frac{x^3}{x-4}|$, We call $f_+ = |\frac{x^3}{x-4}|$ for $x>0$ and $f_- = |\frac{x^3}{x-4}|$ for $x<0$ (and both are $0$ on the rest). We have $f = f_+ + f_-$ for $x<4$.

*

*For $x < 4$, $|x-4| = 4-x$,

*$|x^3| = x^3$ for $x>0$ and $|x^3| = -x^3$ for $x<0$.

*Therefore : $|\frac{x^3}{x-4}| = \frac{x^3}{4-x}$ for $x >0$ and $|\frac{x^3}{x-4}| = \frac{-x^3}{4-x}$ for $x<0$.

*Hence, $f_+ = \frac{x^3}{4-x}$ and $f_- = -\frac{x^3}{4-x}$
Notice that $4-x > 1$ and therefore $f_+ = x^3\cdot\frac{1}{4-x} < x^3 \cdot 1$ for $x>0$ and $f_- = -x^3\cdot\frac{1}{4-x} < -x^3 \cdot 1$
We therefore have a bound $f < |x^3|$ for all $x$, because the signs just flip and cancel for $x>4$ and the cases are analoguous.
You should have no issue fixing $\delta$ now. But here it is, just in case :

 Set $\delta = \varepsilon^{1/3}$

