The derivative of function f(x ), where f(x) ={cases} { x²}{x}, & x ≠ 0 \\ 0, & x=0 {cases}

Correct Answer is : exists at x = 0 and is also continuous

exists at x = 0 but is not continuous

Correct Answer is : exists at x = 0 and is also continuous

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HCU MCA 2013 Paper

Question : 26 of 75

Marks: +1, -0

\begin{cases} \frac{\sin x^{2}}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}

Solution:

(b) Given,

f(x) =

\begin{cases} \frac{\sin x^{2}}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}

Now, we check the continuity of f(x) at x = 0, f(0) = 0,

LHL

\; = f(0-0) = \lim\limits_{h\rightarrow 0} f(0-h) = \lim\limits_{h=0} \frac{\sin h^{2}}{-h}

\; = -\lim\limits_{h\rightarrow 0} \frac{\sin h^{2}}{h^{2}} \times h

\; = -\lim\limits_{h\rightarrow 0} \frac{\sin h^{2}}{h^{2}} \times \lim\limits_{h\rightarrow 0} h

\; = -(1) \times 0 = 0

RHL

\; = f(0+0) = \lim\limits_{h\rightarrow 0} (0+h) = \lim\limits_{h\rightarrow 0} \frac{\sin h^{2}}{h}

\; = \lim\limits_{h\rightarrow 0} \frac{\sin h^{2}}{h^{2}} \times \lim\limits_{h\rightarrow 0} h = 1 \times 0 = 0

∵

\; f(0) = LHL = RHL = 0

Hence, f(x) exists at x = 0, and is also continuous.

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