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Consider $f(x)= x+x^2sin(π/x)$

Does this monotonic in some neighborhood of zero?

I think it isn't as $x^2sin {pi/2}$ oscillates near zero. Even though amplitude of oscillation decreases,the function cannot be monotone.

But how can i prove this?

If $f^{prime}(x)$ takes both positive and negative values in every neighborhood of $0$, then $f$ cannot be monotonic in any neighborhood of $0$.

Now take the derivative of $f(x)$ and observe what happens in a neighbourhood of $0$.

You have

$$f^prime (x) = 1 + 2xsinleft(frac{pi}{x}right) - picosleft(frac{pi}{x}right).$$

Hence for $vert x vert le 1/2$

$$leftvert f^prime(x) - left(1 - picosleft(frac{pi}{x}right)right) rightvert le 1$$

which is equivalent to
$$-pi cos(frac{1}{x}) le f^prime(x)le 2-pi cos(frac{1}{x})$$

Consequently for $n ge 1$ integer:

• when $displaystyle frac{1}{x} in (2npi + frac{2pi}{3},2npi +
  frac{4pi}{3})$ we have $displaystyle cos(frac{1}{x}) le
  -frac{1}{2}$. Hence for $displaystyle x in (frac{1}{2npi + frac{4pi}{3}},frac{1}{2npi + frac{2pi}{3}})$: $f^prime(x)
  ge -pi cos(frac{1}{x}) ge pi/2$. Which implies that $f$ is
  strictly increasing on each interval $(frac{1}{2npi +
  frac{4pi}{3}},frac{1}{2npi + frac{2pi}{3}})$.




• Similarly for $displaystyle frac{1}{x} in ((2n+1)pi +
  frac{5pi}{6},(2n+1)pi + frac{7pi}{6})$ we have $displaystyle
  cos(frac{1}{x}) ge frac{sqrt{3}}{2}$ hence for $displaystyle
  x in (frac{1}{(2n+1)pi + frac{7pi}{6}},frac{1}{(2n+1)pi +
  frac{5pi}{6}})$: $f^prime(x) le 2-pi cos(frac{1}{x}) le
  2-pisqrt{3}/2 <0$. Which implies that $f$ is strictly decreasing
  on each interval $(frac{1}{(2n+1)pi +
  frac{7pi}{6}},frac{1}{(2n+1)pi + frac{5pi}{6}})$.

Finally $f$ is not strictly increasing in any neighborhood of $0$.

However $f^prime (0) =1$. Interesting case!

The 1st-order derivative shows (almost!) everything$$f'(x)=1-pi cos{pi over x}$$which has the roots$$1over 2k+{1overpi }cos^{-1}{{1overpi }}$$and therefore changes its sign permanently around zero. Then the function can't monotonic. Here is a sketch of the derivative

We need not make use of differentiability (or even continuity) of $f$:

Note that for $ninBbb N$,
$$begin{align}fleft(frac1{n-frac12}right)-fleft(frac1{n+frac12}right)&=frac1{n-frac12}+frac 1{(n-frac12)^2}sin (n-tfrac12)pi-frac1{n+frac12}-frac1{(n+frac12)^2}sin(n+tfrac12)pi\
&=frac1{n-frac12}-frac1{n+frac12}-(-1)^ncdotleft(frac 1{(n-frac12)^2}+frac 1{(n+frac12)^2}right)\
&=frac1{n^2-frac14}-(-1)^nfrac{(n+frac12)^2+(n-frac12)^2}{(n^2-frac14)^2}
\&=frac1{n^2-frac14}-(-1)^ncdot2cdot frac{n^2+frac14}{(n^2-frac14)^2}.
end{align}$$
As
$$ frac{n^2+frac14}{(n^2-frac14)^2}>frac{n^2-frac14}{(n^2-frac14)^2}=frac1{n^2-frac14},$$
it follows that $fleft(frac1{n-frac12}right)-fleft(frac1{n+frac12}right)$ is positive for odd $n$ and negative for even $n$. Any interval that has $0$ as interior point also contains $frac1{npmfrac12}$, so that $f$ cannot be monotonic in such an interval. (A similar result holds for $fleft(-frac1{n+frac12}right)-fleft(-frac1{n-frac12}right)$).

Hint :

The derivative of $f(x)$ is :

$$f'(x) = 1 + 2xsinleft(frac{pi}{x}right) - picosleft(frac{pi}{x}right)$$

Now study what happens in a neighborhood of $0$.

Recall that in order to have a monotonicity, it should be $f'(x) >0 forall x in D(0)$ or $f'(x) <0 forall x in D(0)$.