Show that the limit of $ f(x)$ as $x\rightarrow 0$ is $0$

Question

Show directly from the definition of a limit of a function that

lim x->0 (x^(1/3) * sin(1/x)) = 0. 

The definition is

The limit of f as x goes to p is q if for every e>0 there exists a d>0

s.t.

dx(x,p)<d for x belongs to E -> dy(f(x), q)< e

with the notation

limi x->p f(x) = q

My approach

We want to show that

dx(x,0)<d, x belongs to E -> dy( (x^(1/3) * sin(1/x)),0) < e

which is

| x-0| < d - > | x |<d

and

|(x^(1/3) * sin(1/x))|< e 

Now the solution I have takes e^3=d and shows that

|(x^(1/3) * sin(1/x))|< d^(1/3)=e 

which makes no sense...

I suppose the steps start from

|x^(1/3)|<= d^(1/3)

|x^(1/3)| * | sin(1/x)|<= d^(1/3) * | 1| (Does he multiply by one because the max value of sin is 1 and hence it will always be smaller? )

Then takes

e^3=d 

and somehow it is proven. It makes much more sense that I explained it now but I am looking to see what people think....

Please edit the notation if you know how. Thanks a lot.

Answer 1

We want to show that for every positive $\epsilon$, there is a positive $\delta$ such that $$\left|x^{1/3}\sin(1/x)-0\right|\lt \epsilon\quad\text{whenever}\quad 0\lt |x-0|\lt \delta. \tag{1}$$

So let $\epsilon\gt 0$ be given. Note that if $x\ne 0$ then $$\left|x^{1/3}\sin(1/x)-0\right|\le |x|^{1/3},\tag{1}$$ since $|\sin(1/x)|\le 1$.

If $|x|\lt \epsilon^3$, then $|x|^{1/3}\lt \epsilon$. Let $\delta=\epsilon^3$. Then whenever $0\lt |x|\lt \delta$, by (2) we will have $\left|x^{1/3}\sin(1/x)-0\right|\lt \epsilon$. We have shown that for every $\epsilon\gt 0$, there is a $\delta\gt 0$ such that (1) holds. This completes the proof.