Let $f:[−1, 1] \to \mathbb R$ be defined by $f(x) = \frac{x^2+ [\sin\pi x]}{1+|x|}$

Question

$[y]$ denotes the greatest integer less than or equal to $y$

Let $f:[−1, 1] \to \mathbb{R}$ be defined by $f(x) = \dfrac{x^2+ [\sin\pi x]}{1+|x|}$.

I don't understand why this function is discontinuous at $x=0$ and $x=\frac{1}{2}$

Answer 1

Hint: for $n \in \mathbb N$ we have

$f(-1/n) \to -1 \ne 0=f(0)$ as $ n \to \infty$.

Answer 2

$$\frac{x^2}{1+|x|} + \frac{[sin \pi x]}{1+|x|}$$

The left hand side of the addition we don't have to worry about because the function is continuous.(trivial to see!) However the step function is not continuous so we need to analyse the right hand side function.

Let, $$L^- = lim_{x \to 0^{-}} \frac{[sin \pi x]}{1+|x|}$$ $$L^+ = lim_{x \to 0^{+}} \frac{[sin \pi x]}{1+|x|}$$

At $x \to 0^-$, $sin(\pi x)$ is slightly less than $0$, therefore $\lim_{x \to 0^{-}} [sin(\pi x)]$ will go to $-1$.
For $x \to 0^{+}, sin(\pi x)$ is slightly bigger than $0$, therefore $\lim_{x \to 0^{+}} [sin(\pi x)]$ will go to $0$.

And therefore $L^{+} = 0$ and $L^{-} = -1$.

It will help you if you draw the curve of $sin(\pi x)$ and visualize the tending to from the right and the left for any particular $x$ to see how $sin(\pi x)$ changes value.

Hopefully you can do the same for $x = 1/2$.

Cheers!

Answer 3

For $0<x<1/2$, you have $$ [\sin(\pi x)]=0 $$ while $[\sin(\pi/2)]=1$. Compute $$ \lim_{x\to 1/2^-}f(x)=\lim_{x\to 1/2^-}\frac{x^2}{1+|x|} $$ and $f(1/2)$.

What about $x=0$?