Question
$x = 0$ पर फलन $f(x) = |x|$ है
$f(x) = \,|\,\,x\,\,|\,\, = \,\,|0|\,\, = 0$ व $f(0 + h) = f(h) = \,\,|h|$
$\therefore \,\,\mathop {\lim }\limits_{h \to 0 - } \,\frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0 - } \,\frac{{|h|}}{h} = - 1$
एवं $\mathop {\lim }\limits_{h \to 0 + } \,\frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0 + } \,\frac{{|h|}}{h} = 1$
अत: यह सतत् एवं अन्अवकलनीय है।
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$\left[\frac{x}{\sqrt{x^2-y^2}}+e^{\frac{y}{x}}\right] x \frac{d y}{d x}=x+\left[\frac{x}{\sqrt{x^2-y^2}}+e^{\frac{y}{x}}\right] y$
का हल वक्र $y = y ( x )$ हो जो $(1,0)$ तथा $(2 \alpha, \alpha), \alpha > 0$ से होकर गुजरता हो तब $\alpha$ बराबर होगा