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Question Number 413 by 123456 last updated on 25/Jan/15
lim_((x,y)→(0,0)) (((x+y)(x^2 −y^2 ))/( (√(x^2 +y^2 ))))
$$\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\:\:\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }} \\ $$
Answered by prakash jain last updated on 31/Dec/14
x=r cos t, y=r sin t f=((r(cos t+sin t)r^2 (cos^2 t−sin^2 t))/r) f=r^2 (cos t+sin t)(cos^2 t−sin^2 t) lim_(r→0) f=0
$${x}={r}\:\mathrm{cos}\:{t},\:{y}={r}\:\mathrm{sin}\:{t} \\ $$$${f}=\frac{{r}\left(\mathrm{cos}\:{t}+\mathrm{sin}\:{t}\right){r}^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} {t}−\mathrm{sin}^{\mathrm{2}} {t}\right)}{{r}}\: \\ $$$${f}={r}^{\mathrm{2}} \left(\mathrm{cos}\:{t}+\mathrm{sin}\:{t}\right)\left(\mathrm{cos}^{\mathrm{2}} {t}−\mathrm{sin}^{\mathrm{2}} {t}\right) \\ $$$$\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}=\mathrm{0} \\ $$

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