# Determine the discontinuities of the following function

Determine the discontinuities of the following function:

$$f(x) = \cos \left(\frac{1}{x}\right) , \ \ x<0\\$$ $$f(x)= x^2, \ \ 0 \leq x \leq 1$$ $$f(x)=\frac{x^2-3x+2}{x-2} , \ \ 1<x<2$$ $$f(x) =2 , \ \ x=2$$ $$f(x) =\sqrt{x-1} , \ \ x>2$$

For each discontinuity identified above, classify whether the discontinuity are removable, jump discontinuity, infinite or oscillating.

From the definition of the function, we see that;

The functions $\ f(x) \$ has discontinuities at $\ x=0, \ x=1, \ x=2 \$

Among this,

$x=2 \$ is removable singularity, jump discontinuity.

$x=1 \$ is jump discontinuity.

$x=0 \$ is infinite discontinuity.

Does this seem to be correct?

• Yes, it seems so – charmd Jun 5 '18 at 6:04
• Your answer for $x=0$ is wrong. It is an oscillating discontinuity since $\cos (\frac 1 x)$ oscillates as $x \to 0-$. – Kabo Murphy Jun 5 '18 at 6:17

Hints:

• $f(0^-)$ does not exist, as the function is oscillating.

• $f(0)=f(0^+)=0$.

Hence oscillating.

• $f(1^-)=f(0)=1$.

• $f(1^+)=0$.

Hence jump.

• $f(2^-)=1$ (you can simplify by $x-2$).

• $f(2)=2$.

• $f(2^+)=1$.

Hence removable. At $x=0$, it is oscillatory discontinuity but not infinite discontinuity. It is oscillatory discontinuity because here, for$\ \ x<0 , \cos \left(\frac{1}{x}\right)$ appear to be approaching many values simultaneously as $x \to 0^-$.

Infinite discontinuity means the function goes to infinity at that point. Both Left Hand Limit and Right Hand Limit are not finite.

At $x=1$, it is jump discontinuity. The function is approaching different values depending on the direction x is coming from. As $x \to 1^-$, $f(x) \to 1$ and $x \to 1^+$, $f(x) \to 0$.

At $x=2$, it is a removable singularity but not jump discontinuity. At jump discontinuity, LHL $\neq$ RHL. But here they are equal.