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Question: Which of the following function(s) not defined at $x = 0$ has/have removal discontinuity at $x = 0$ ...

Which of the following function(s) not defined at x=0x = 0 has/have removal discontinuity at x=0x = 0 ?

A

f(x)=11+2cot⁡xf(x) = \frac{1}{1+2^{\cot x}}

B

f(x)=cos⁡(∣sin⁡x∣x)f(x) = \cos (\frac{|\sin x|}{x})

C

f(x)=xsin⁡πxf(x) = x \sin \frac{\pi}{x}

D

f(x)=1ln⁡∣x∣f(x) = \frac{1}{\ln|x|}

Answer

f(x) = cos⁡(∣sin⁡x∣x\cos (\frac{|\sin x|}{x}), f(x) = xsin⁡πxx \sin \frac{\pi}{x}, f(x) = 1ln⁡∣x∣\frac{1}{\ln|x|}

Explanation

Solution

A function f(x)f(x) has a removable discontinuity at x=ax=a if lim⁡x→af(x)\lim_{x\to a} f(x) exists and is finite, but is not equal to f(a)f(a) (or f(a)f(a) is undefined). Since the problem states the functions are not defined at x=0x=0, we only need to check if the limit lim⁡x→0f(x)\lim_{x\to 0} f(x) exists and is finite. This requires the left-hand limit (LHL) and the right-hand limit (RHL) at x=0x=0 to be equal and finite.

  1. f(x)=11+2cot⁡xf(x) = \frac{1}{1+2^{\cot x}}

    • LHL: As x→0−x \to 0^-, cot⁡x→−∞\cot x \to -\infty. lim⁡x→0−f(x)=11+2−∞=11+0=1\lim_{x\to 0^-} f(x) = \frac{1}{1+2^{-\infty}} = \frac{1}{1+0} = 1.
    • RHL: As x→0+x \to 0^+, cot⁡x→+∞\cot x \to +\infty. lim⁡x→0+f(x)=11+2∞=11+∞=0\lim_{x\to 0^+} f(x) = \frac{1}{1+2^{\infty}} = \frac{1}{1+\infty} = 0.
    • LHL ≠\neq RHL. This is a non-removable discontinuity.

  2. f(x)=cos⁡(∣sin⁡x∣x)f(x) = \cos \left(\frac{|\sin x|}{x}\right)

    • LHL: As x→0−x \to 0^-, ∣sin⁡x∣x=−sin⁡xx→−1\frac{|\sin x|}{x} = \frac{-\sin x}{x} \to -1. lim⁡x→0−f(x)=cos⁡(−1)=cos⁡(1)\lim_{x\to 0^-} f(x) = \cos(-1) = \cos(1).
    • RHL: As x→0+x \to 0^+, ∣sin⁡x∣x=sin⁡xx→1\frac{|\sin x|}{x} = \frac{\sin x}{x} \to 1. lim⁡x→0+f(x)=cos⁡(1)\lim_{x\to 0^+} f(x) = \cos(1).
    • LHL = RHL = cos⁡(1)\cos(1). The limit exists and is finite. This is a removable discontinuity.

  3. f(x)=xsin⁡πxf(x) = x \sin \frac{\pi}{x}

    • LHL: As x→0−x \to 0^-, x→0x \to 0 and sin⁡πx\sin \frac{\pi}{x} is bounded between -1 and 1. By the Squeeze Theorem, lim⁡x→0−xsin⁡πx=0\lim_{x\to 0^-} x \sin \frac{\pi}{x} = 0.
    • RHL: As x→0+x \to 0^+, x→0x \to 0 and sin⁡πx\sin \frac{\pi}{x} is bounded between -1 and 1. By the Squeeze Theorem, lim⁡x→0+xsin⁡πx=0\lim_{x\to 0^+} x \sin \frac{\pi}{x} = 0.
    • LHL = RHL = 0. The limit exists and is finite. This is a removable discontinuity.

  4. f(x)=1ln⁡∣x∣f(x) = \frac{1}{\ln|x|}

    • LHL: As x→0−x \to 0^-, ∣x∣→0+|x| \to 0^+, so ln⁡∣x∣→−∞\ln|x| \to -\infty. lim⁡x→0−f(x)=1−∞=0\lim_{x\to 0^-} f(x) = \frac{1}{-\infty} = 0.
    • RHL: As x→0+x \to 0^+, ∣x∣→0+|x| \to 0^+, so ln⁡∣x∣→−∞\ln|x| \to -\infty. lim⁡x→0+f(x)=1−∞=0\lim_{x\to 0^+} f(x) = \frac{1}{-\infty} = 0.
    • LHL = RHL = 0. The limit exists and is finite. This is a removable discontinuity.

The functions with removable discontinuities at x=0x=0 are cos⁡(∣sin⁡x∣x)\cos \left(\frac{|\sin x|}{x}\right), xsin⁡πxx \sin \frac{\pi}{x}, and 1ln⁡∣x∣\frac{1}{\ln|x|}.