[College Calc 01-03. Limits of Functions] #04

We only need observe the point $x=0.$

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A. True

Suppose $p< 0.$ We most certainly have that $\displaystyle \lim_{n\to\infty}f\left(\frac{2}{(2n+1)\pi}\right) = \infty,$ such that $\displaystyle \lim_{x\to 0}f(x)$ can never converge.

Similarly if $p=0,$ then $\displaystyle \lim_{n\to\infty}f\left(\frac{2}{(2n+1)\pi}\right) = 0$ and $\displaystyle \lim_{n\to\infty}f\left(\frac{1}{n\pi}\right) = 0,$ such that $\displaystyle \lim_{x\to 0}f(x)$ can never converge.

However given $p>0,$ we have $-|x|^p \leq f(x) \leq |x|^p$ such that $\displaystyle \lim_{x\to 0} f(x) = 0$ by the squeeze theorem .

Therefore such a minimal integer $p$ is $1.$

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B. True

Differentiability at $x=0$ is equivalent to the statement that the limit

$\lim_{x\to 0}\frac{f(x)-f(0)}{x}$

converges. Let us see if it does.

$\lim_{x\to 0}\frac{f(x)-f(0)}{x} = \lim_{x\to 0}\frac{x^p\sin\frac{1}{x}}{x} = \lim_{x\to 0} x^{p-1}\sin\frac{1}{x}.$

As with the solution for A , the equivalent condition for this limit to converge is $p>1.$

Therefore such a minimal integer $p$ is $2.$

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C. True

First let us differentiate $f(x).$ For it to be defined at $x=0$ we must already have that $p>1.$

$f'(x) = \begin{cases}0 & \textrm{if } x = 0; \\ px^{p-1}\sin\dfrac{1}{x}-x^{p-2}\cos\dfrac{1}{x} & \textrm{otherwise.}\end{cases}$

As with the solution for A , the equivalent condition for $f'(x)$ to be continuous is $p>2.$

Therefore such a minimal integer $p$ is $3.$

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The moral of this problem is that differentiability is not equivalent to the continuity of the derivative. Although they may seem equivalent in most cases that you would come across in basic calculus problems, you must still be aware that they are not.