Prove Function is Differentiable

[SCGirl]

Hi, here is the question I am stuck on:

Use the definition of "f is differentiable at a" to show that the following piecewise function:

. . . . . . . .. ./ x^2 sin(1/x) for x not equal to 0
. . .g(x) = <
. . . . . . . .. .\ . . .0 . . . . . . for x = 0

...is differentiable at 0.

Thanks in advance.
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Edited by stapel -- Reason for edit: formatting, clarity

 

[stapel]

How far have you gotten in the process of taking the limit of x² sin(1/x) as x tends toward zero?

Please reply showing all of your work and reasoning so far. Thank you.

Eliz.

 

[SCGirl]

Ok, I have cancelled out the x. So I got x times sin(1), so it equals zero, I think.

 

[stapel]

Ok, I have cancelled out the x. So I got x times sin(1), so it equals zero, I think.

What...?!? :shock:

Are you actually saying that you "cancelled" an "x" from the squaring with the "x" in the denominator of the argument of the sine function...? Surely I'm not understanding you correctly!

Please reply with clarification, showing all of your steps. You set up the limit of x² sin(1/x) as x tends to zero, and... then what? How did you use the limit you were given for [sin(x)] / x ?

Thank you.

Eliz.

 

[pka]

\(\displaystyle \L \begin{array}{rcl}
f'(x_0 ) = \lim _{h \to 0} \frac{{f(x_0 + h) - f(x_0 )}}{h} \\
x_0 = 0\quad & \Rightarrow & \quad \lim _{h \to 0} \frac{{\left( {0 + h} \right)^2 \sin \left( {\frac{1}{{0 + h}}} \right) - f(0)}}{h} \\
& = & \lim _{h \to 0} \frac{{h^2 \sin \left( {\frac{1}{h}} \right) - 0}}{h} \\
& = & \lim _{h \to 0} h \sin \left( {\frac{1}{h}} \right) \\
& = & 0 \\
\end{array}\)
Because sine is bounded and h approaches 0.

 

[SCGirl]

Thank you, pka. That makes sense.