How to Prove that this Function is not $1-1$ I'm trying to show that the function 

$$f(x)= \dfrac{x}{4}+x^2\sin\left(\dfrac{1}{x}\right)$$
is not $1-1$ for any neighborhood of $0$. I know that what I have to do is find two different points that yield the same value, but these are not so obvious.

Thanks for any help.
The derivative of $f(x)$ is

$$f'(x)=\frac 14+2x\sin\left(\frac 1x\right)-\cos\left(\frac 1x\right)$$

In any neighborhood of zero, you can find a very small positive $a$ where $\cos\left(\frac 1a\right)=1$. Then the second term of my expression for $f'(a)$ is near zero, thus $f'(a)$ is near $-\frac 34$, thus $f'(a)$ is negative, thus $f(x)$ is decreasing in a neighborhood of $a$ that is inside our neighborhood of zero.

In that same neighborhood of zero, you can find a very small positive $b$ where $\cos\left(\frac 1b\right)=-1$. Then the second term of my expression for $f'(b)$ is near zero, thus $f'(b)$ is near $\frac 54$, thus $f'(b)$ is positive, thus $f(x)$ is decreasing in a neighborhood of $b$ that is inside our neighborhood of zero.

Both $a$ and $b$ are positive, so $f(x)$ is continuous on the interval between $a$ and $b$. You can easily show that any function continuous on an interval that increases in a subinterval and also decreases in a subinterval is not injective on that interval. (Use the Intermediate Value Theorem on the four cases $a&lt;b$ or $b&lt;a$, $f(a)&lt;f(b)$ or $f(b)&lt;f(a)$). This is a useful theorem to use in many other problems, as well.

Thus your function $f(x)$ is not injective in any neighborhood of zero, in the positive part of that interval alone.
When $-\frac 1n&lt;x&lt;\frac 1n$ the argument of $\sin (\frac 1x)$ takes all the values of the interval $]n,\infty[$ so that $\sin (\frac 1x)$ takes infinitely many times all the values of the rank $[-1,1]$ of the function $\sin x$ (If you want to draw the graph of $\sin (\frac 1x)$, you have a shaded rectangle in the neighborhood of $0$).

This is the reason for which the function $g(x)=x^2\sin (\frac 1x)$ can't be 1−1 for any neighborhood of $0$ and so is for $f(x)=\frac x4 +x^2\sin(\frac 1x)$.