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I was asked a question for homework to find the derivative of the following function:

$$f(x) =\begin{cases} x^3(1+\sin \left(\frac{1}{x})\right)&\text{if}\; x \neq 0\\ \\ 0&\text{if}\;x=0 \end{cases}$$

I have started off by stating epsilon-delta definition of a limit by stating that for all $\varepsilon >0$, there exists a $\delta >0$ such that if $$0<|x-a|< \delta, \;\text{then}\; |f(x) - L|< \varepsilon$$

I have then stated that since the limit of the function is $0$ (By first principles) then $$\lim_{h \to0} \frac{f(0+h)-f(0)}{h} =\lim_{h \to 0} \frac{h^3(1+\sin (1/h))}{h}$$

This then can simplify down and by substituting the function into the definition for when the limit is $0$ we get, $|h^2(1+\sin(1/x)) - 0|< \varepsilon$.

And this is where I get stuck. I'm not too sure how to find the solution from this point onwards. Any help would be greatly appreciated. Thanks in advance.

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  • $\begingroup$ Just aside: refer this for $\LaTeX$ symbols $\endgroup$
    – Chinnapparaj R
    Apr 7, 2019 at 11:17
  • $\begingroup$ Do you know the result stating that the product of a function with limit $0$ and a bounded function has limit zero? $\endgroup$
    – user539887
    Apr 7, 2019 at 11:20

2 Answers 2

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I am not sure what you know yet and what you don't, but consider the rule that states

Let $f(x)$ and $g(x)$ be functions s.t $\lim(f(x)) = 0$ and $|g(x)|<M$ for M>0, then $\lim(f(x)*g(x)) = 0$.

If you are allowed to use this rule then you have done all the work already! Why? If you aren't sure (try by yourself first):

$ -1 <= \sin(x) <= 1 $ and so $ 0<=1-\sin(1/h)<=2 $ so $|1-\sin(1/h)|<=2$

In our case $g(x) = 1-\sin(1/h)$ and $f(x) = h^2$

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  • $\begingroup$ I think that we need to prove that it is differentiable first before we can apply limit laws $\endgroup$
    – JayP
    Apr 7, 2019 at 11:58
  • $\begingroup$ Well that is what you are trying to prove isn't it? To show a function is differentiable in a specific point you need to show the limit of $limh→0f(0+h)−f(0)h$ exists, and when you find its value - it is the derivative at that point. Once you have already started calculating the limit of that formula, you can look at it as a different function that you need to find the limit of. You know the limit of $x$ when $x -> 0$ is zero, same goes for h. and you know the function its multiplied by is bounded by 2, so why wouldn't you be able to use the rule? $\endgroup$
    – ash
    Apr 8, 2019 at 21:01
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$$\left|\frac{f(x)}{x}\right|\le x^2\big(1+\sin{(1/x)}\big)\le 2x^2\longrightarrow_0 0$$

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