Question

Check whether the following functions are continuous or not at x = 0. Also, find
the nature of discontinuity at that point, if it exists.
(i) f(x)= {[(2-x)^(1/2) - (2+x)^(1/2)] / [x] , x≠0 ; 1/√2 , x=0
(ii) f(x)= { x^(2)+1/3, x≤0 ; -[x^(3)+1/3], x>0

Answer 1

Answer on Question #69038 – Math – Real Analysis

Question

Check whether the following functions are continuous or not at $x = 0$. Also, find the nature of discontinuity at that point, if it exists.

(i) $f(x) = \left\{ \left[ (2 - x)^{(1/2)} - (2 + x)^{(1/2)} \right] / [x] \right\}, x \neq 0; 1/\sqrt{2}, x = 0$

(ii) $f(x) = \left\{ x^{(2)} + 1/3, x \leq 0; -[x^{(3)} + 1/3], x > 0 \right.$

Solution

(i) $f(x) = \begin{cases} \dfrac{\sqrt{2 - x} - \sqrt{2 + x}}{x}, & x \neq 0 \\ \dfrac{1}{\sqrt{2}}, & x = 0 \end{cases}$

$$\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0} f(x) = -\dfrac{1}{\sqrt{2}}.$$ $$f(0) = \dfrac{1}{\sqrt{2}}.$$

Thus, $f(x)$ has a removable discontinuity at $x = 0$.

(ii) $f(x) = \begin{cases} x^2 + \dfrac{1}{3}, & x \leq 0 \\ -x^3 - \dfrac{1}{3}, & x > 0 \end{cases}$

$$\lim_{x \to 0^{-}} f(x) = \dfrac{1}{3}, \quad \lim_{x \to 0^{+}} f(x) = -\dfrac{1}{3}.$$

Thus, $f(x)$ has a jump discontinuity at $x = 0$ (discontinuity of the first kind).

Answer: (i) discontinuous; a removable discontinuity; (ii) discontinuous, a jump discontinuity.

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