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Exercise 8.3 · Q5

Q.Let g(x,y)=x2yx4+y2g(x,y)=\dfrac{x^2y}{x^4+y^2} for (x,y)≠(0,0)(x,y)\ne(0,0) and f(0,0)=0f(0,0)=0.

(i) Show that lim⁡(x,y)→(0,0)g(x,y)=0\displaystyle\lim_{(x,y)\to(0,0)}g(x,y)=0 along every line y=mx, m∈Ry=mx,\ m\in\mathbb R.
(ii) Show that lim⁡(x,y)→(0,0)g(x,y)=k1+k2\displaystyle\lim_{(x,y)\to(0,0)}g(x,y)=\dfrac{k}{1+k^2} along every parabola y=kx2, k∈R∖{0}y=kx^2,\ k\in\mathbb R\setminus\{0\}.
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Substitute each path directly into gg, simplify algebraically (cancelling the common power of xx), and then let x→0x\to0.

Part (i): along y=mxy=mx.

g(x,mx)=x2(mx)x4+(mx)2=mx3x4+m2x2=mx3x2(x2+m2)=mxx2+m2(x≠0).g(x,mx) = \frac{x^2(mx)}{x^4+(mx)^2} = \frac{mx^3}{x^4+m^2x^2} = \frac{mx^3}{x^2(x^2+m^2)} = \frac{mx}{x^2+m^2} \qquad (x\ne0).

As x→0x\to0: numerator →0\to0, denominator →m2\to m^2 (a fixed nonzero constant, since m∈Rm\in\mathbb R is fixed for this path — if m=0m=0 then g≡0g\equiv0 trivially). So lim⁡x→0mxx2+m2=0m2=0\displaystyle\lim_{x\to0}\frac{mx}{x^2+m^2}=\frac{0}{m^2}=0 for every mm.

∴ lim⁡(x,y)→(0,0)g(x,y)=0 along every line y=mx.\therefore\ \lim_{(x,y)\to(0,0)}g(x,y)=0 \text{ along every line } y=mx.

Part (ii): along y=kx2, k≠0y=kx^2,\ k\ne0.

g(x,kx2)=x2(kx2)x4+(kx2)2=kx4x4+k2x4=kx4x4(1+k2)=k1+k2(x≠0).g(x,kx^2) = \frac{x^2(kx^2)}{x^4+(kx^2)^2} = \frac{kx^4}{x^4+k^2x^4} = \frac{kx^4}{x^4(1+k^2)} = \frac{k}{1+k^2} \qquad (x\ne0).

This ratio is constant — it doesn't depend on xx at all — so as x→0x\to0,

lim⁡x→0g(x,kx2)=k1+k2.\lim_{x\to0}g(x,kx^2) = \frac{k}{1+k^2}. …

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