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Which of the following functions are not defined at $x=0$ / have removable discontinuity at the origin

(a) $f(x) =\frac {1}{1+2^{\cot x}}$

(b) $f(x) =\cos \frac {|\sin x|}{x}$

(c) $f(x) = x \sin \left(\frac\pi x\right)$

(d) $f(x) = \frac{1}{\log|x|}$

According to me (a) and (d) should be the answer as it is not defined at $x=0$ but this doesn't match the answer. What is the way to check removable discontinuty

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Aladdin
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  • Can you write down the definition of "removable discontinuity"? Because from what you wrote, in particular "as it is not defined at x=0", you don't seem to have an understanding about what removable discontinuities are. – 5xum Aug 20 '18 at 12:47
  • Well according to me removable discontunity is due to. Function not defined at a point but we can introduce some value to make it continious. – Aladdin Aug 20 '18 at 12:49
  • OK. Well none of the functions are defined at $x=0$... – 5xum Aug 20 '18 at 12:49
  • Every single one of these functions is not defined in $0$. – giobrach Aug 20 '18 at 12:50
  • Why is (b) and (c) not defined at x=0.Their limits are equal. – Aladdin Aug 20 '18 at 12:52
  • @GENESECT (c) is not defined at $x=0$ because $\frac\pi x$ is not defined for $x=0$. Therefore, the sine of that value is not defined, because $\sin(y)$ is not defined if $y$ is defined... – 5xum Aug 20 '18 at 12:57

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Hint:

Try to look at the value of $$\lim_{x\to 0} f(x)$$

How do you think this limit is connected to removable discontinuities?

5xum
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  • Well I think limit should exist for removable discontinuity – Aladdin Aug 20 '18 at 12:56
  • @GENESECT There you go. Now you have a method for solving your problem. – 5xum Aug 20 '18 at 12:57
  • What can we say for function... Is it necessary for it to exist. Like (c) the function doesn't exist but limit exists – Aladdin Aug 20 '18 at 13:01
  • @GENESECT The limit at $0$ exists, but the function is not defined at $0$ if and only if the function has a removable discontinuity at $0$. – 5xum Aug 20 '18 at 13:02