[Solved] What is Derivative of x^2 sin(1/x) at x=0?

Answer: The derivative of x^2 sin(1/x) at x=0 is equal to 0 if the function is defined to be 0 at x=0.

Question

Find the derivative of the function

f(x) = $\begin{cases} x^2 \sin(\frac{1}{x}) & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$

at x=0.

That is, find f'(0).

Solution

From definition, the first order derivative of f(x) at x=0 is given by the following limit

f'(0) = lim_h→0 $\dfrac{f(0+h)-f(0)}{h}$

= lim_h→0 $\dfrac{f(h)-f(0)}{h}$

= lim_h→0 $\dfrac{f(h)-f(0)}{h}$

= lim_h→0 $\dfrac{h^2 \sin(\frac{1}{h})-0}{h}$

= lim_h→0 $h \sin(\frac{1}{h})$

So we obtain that

f'(0) = lim_h→0 $h \sin(\frac{1}{h})$ ...(∗)

Now lets calculate the limit lim_h→0 $h \sin(\frac{1}{h})$. We will use the sandwich/squeeze theorem of limits. It is known that

-1 ≤ $\sin(\frac{1}{h})$ ≤ 1 …(∗∗)

For h>0, multiplying both sides of (∗∗) by h we get that -h ≤ $h\sin(\frac{1}{h})$ ≤ h. Take limit h→0+ on both sides, $\lim\limits_{h \to 0+} (-h)$ ≤ $\lim\limits_{h \to 0+} h\sin(\frac{1}{h})$ ≤ $\lim\limits_{h \to 0+} h$ ⇒ 0 ≤ $\lim\limits_{h \to 0+} h\sin(\frac{1}{h})$ ≤ 0 ∴ limh→0+ $h\sin(\frac{1}{h})$ = 0 …(A)

On the other hand, in the same way as above, we have lim_h→0- $h\sin(\frac{1}{h})$ = 0. The calculation is shown below.

When h<0, multiplying (∗∗) by h we have h ≤ $h\sin(\frac{1}{h})$ ≤ -h. Taking limit h→0- on both sides, it follows that $\lim\limits_{h \to 0-} h$ ≤ $\lim\limits_{h \to 0-} h\sin(\frac{1}{h})$ ≤ $\lim\limits_{h \to 0-}(-h)$ ⇒ 0 ≤ $\lim\limits_{h \to 0-} h\sin(\frac{1}{h})$ ≤ 0 ∴ limh→0- $h\sin(\frac{1}{h})$ = 0 …(B)

From (A) and (B), we deduce that $\lim\limits_{h \to 0} h\sin(\frac{1}{h})$ = 0. Therefore from (∗), it follows that

f'(0) = 0.

So the derivative of the function f(x) = x²sin(1/x) if x≠0 with f(0)=0 at the point x=0 is equal to 0.

More Derivatives

FAQs