Question

decide if the following function (f(x)) is differentiable at (x = 0). try zooming in on a graphing calculator, or calculating the derivative (f(0)) from the definition.

(f(x)=x^{2}sin(3/x)),

if (x
eq0), and

(f(x) = 0)

if (x = 0).

if it is differentiable, what is the derivative? (if it isnt, enter dne.)

(f(0)=)

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function $y = f(x)$ at $x = a$ is given by $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $a = 0$, $f(0)=0$, and for $h
eq0$, $f(h)=h^{2}\sin(3/h)$. So, $f^{\prime}(0)=\lim_{h
ightarrow0}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0}\frac{h^{2}\sin(3/h)-0}{h}$.

Step2: Simplify the limit expression

Simplify $\lim_{h
ightarrow0}\frac{h^{2}\sin(3/h)-0}{h}$ to $\lim_{h
ightarrow0}h\sin(3/h)$.

Step3: Use the Squeeze - Theorem

We know that $- 1\leqslant\sin(3/h)\leqslant1$ for all $h
eq0$. Multiply each part of the inequality by $h$ (assuming $h
eq0$), we get $-h\leqslant h\sin(3/h)\leqslant h$.
As $h
ightarrow0$, $\lim_{h
ightarrow0}(-h)=0$ and $\lim_{h
ightarrow0}h = 0$. By the Squeeze - Theorem, $\lim_{h
ightarrow0}h\sin(3/h)=0$.

Answer:

$0$