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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.

Answer:

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?

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

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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.

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