Why doesn't MVT hold when I take the limit as $x\to 0$?

Question

I'm considering $f(x) = x^2\sin(1/x)$ for $x\ne 0$, and $f(0)=0$.

MVT gives me $(f(x)-f(0))/ x = x \sin(1/x) = f'(c) = 2c \sin(1/c) - \cos(1/c)$ for some $c$ between $x$ and $0$.

Taking the limit as $x$ goes to $0$ gives me $0 = \lim_{c\to 0} \cos(1/c)$ which doesn't make sense.

What did I do wrong?

Answer 1

This is a classic mistake when applying the mean value theorem. Note that the $c$ obtained from the mean value theorem might be different for each $x$. You apply mean value theorem once for each single $x$, and get a (likely different) $c$ for each single $x$, so you really should write $c_x$ or $c(x)$ to express this dependency.

Now, when taking limits, you cannot treat $c$ as an independent variable without justification, since it does depend on $x$. In other words, even though $\displaystyle\lim_{x\to 0} c(x)=0$, we still could have $$ \lim_{x\to 0} \cos(1/c(x)) \neq \lim_{c\to 0} \cos(1/c). $$ An example is given in my comment, i.e. $c(x)=1/(\pi/2+\lfloor x^{-1} \rfloor\pi)$.

Answer 2

Consider the plots of $\frac{f(x)-f(0)}{x-0}$ and $f'(x)$:

1. $f'(x)$ is continuous and $$f'\!\left(\frac1{(2n-1)\pi}\right)=1\tag{1a}$$ and $$ f'\!\left(\frac1{2n\pi\vphantom{()}}\right)=-1\tag{1b} $$
2. since $|\sin(x)|\le1$, $$ \left|\frac{f(x)-f(0)}{x-0}\right|\le x\tag2 $$

Thus, for any $x$ so that $0\lt x\lt1$, $$ -1\lt\frac{f(x)-f(0)}{x-0}\lt1\tag3 $$ For any $x\gt0$, there is an $n$ so that $\frac1{2n\pi}\lt\frac1{(2n-1)\pi}\le x$.

The Intermediate Value Theorem says that there is a $c$ between $\frac1{2n\pi}$ and $\frac1{(2n-1)\pi}$ so that $f'(c)=\frac{f(x)-f(0)}{x-0}$, which validates the claim of the Mean Value Theorem for this $f$.

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Although for all $x\gt0$, there is at least one $c$ between $0$ and $x$ so that $$ f'(c)=\frac{f(x)-f(0)}{x-0}\tag4 $$ this does not mean that all values of $f'$ are attained.

The right end of the red line traces $\frac{f(x)-f(0)}{x-0}$ and the left end of the red line traces only part of $f'(x)$. Note that any point on the plot of $f'(x)$ on the same horizontal line as and to the left of $\frac{f(x)-f(0)}{x-0}$ (of which there are infinitely many) will serve as $f'(c)$. In the animation, we use the greatest $c$ possible.

Thus, although $\lim\limits_{x\to0}f'(x)$ does not exist, following the left end of the red line, we see that $$ \lim\limits_{x\to0}f'(c(x))=0\tag5 $$