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Question Number 412 by 123456 last updated on 25/Jan/15

$$\underset{(x,y)\to (0,0)}{\lim }\frac{\sin (x^2+y^2)}{\sqrt{x^2+y^2}}$$

Answered by prakash jain last updated on 31/Dec/14

$$f(x,y)=\frac{\sin (x^2+y^2)}{\sqrt{x^2+y^2}}$$$$=\frac{\sin (x^2+y^2)}{x^2+y^2}\sqrt{x^2+y^2}$$$$\mathrm{For}\mathrm{the}\mathrm{case}0<\sqrt{x^2+y^2}<\delta$$$$\mid f(x,y)-0\mid =\mid \frac{\sin (x^2+y^2)}{x^2+y^2}\sqrt{x^2+y^2}\mid$$$$=\mid \frac{\sin (x^2+y^2)}{x^2+y^2}\mid \mid \sqrt{x^2+y^2}\mid <\mid \sqrt{x^2+y^2}\mid <\delta$$$$\underset{(x,y)\to (0,0)}{\lim }f\left(x\right)=0$$

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