← 2016 Paper 1

UPSC 2016 Maths Optional Paper 1 Q3b — Step-by-Step Solution

15 marks · Section A

Functions of two/three variables: limits, continuity · Calculus · asked 4× in 13 yrs · Read the full method →

Question

Let

Find a $\delta>0$ such that $|f(x,y)-f(0,0)|<0.01$, whenever $\sqrt{x^2+y^2}<\delta$.

Typo flag. As printed, the middle term is $5x^2y^2$ (degree 4), which makes the numerator non-homogeneous. Then along $y=kx$ one gets $f\to-\dfrac{5k^2}{(k^2+1)^2}\ne0$ (path-dependent), so $\lim_{(x,y)\to0}f$ does not exist, $f$ is discontinuous at the origin, and no such $\delta$ exists. The intended (and standard) problem has the middle term $5x^2y^3$, giving a homogeneous degree-$5$ numerator; that version is solved below.

Technique

Polar substitution collapses the homogeneous quotient to $f=r\,g(\theta)$, so $|f|\le r\,M$ with $M=\max_\theta|g(\theta)|$; bound $M$ by the triangle inequality and pick $\delta=\varepsilon/M$.

Solution

Strategy. The intended numerator $2x^4y-5x^2y^3+y^5$ is homogeneous of degree $5$ and the denominator $(x^2+y^2)^2$ is homogeneous of degree $4$, so in polar coordinates the quotient is exactly $r$ times a bounded function of $\theta$ alone. Bounding that function gives $|f|\le Mr$, which converts the target $|f|<0.01$ directly into a condition $r<\delta=0.01/M$. We work with $\varepsilon=0.01$ throughout.

Step 1 — Polar substitution

Put $x=r\cos\theta,\;y=r\sin\theta$ with $r=\sqrt{x^2+y^2}>0$. Then $x^2+y^2=r^2$, so the denominator is $(x^2+y^2)^2=r^4$, and the numerator is

Hence, for $(x,y)\ne(0,0)$,

Since $f(0,0)=0$, equation $(1)$ gives the exact identity

Step 2 — Bound the angular factor $g(\theta)$

Using $|\cos\theta|\le 1$ and $|\sin\theta|\le 1$, the triangle inequality gives a uniform bound, independent of $\theta$:

So with $M:=8$ we have $|g(\theta)|\le M$ for every $\theta$. (This safe bound is comfortably enough to score full marks; the true maximum is smaller, but $M=8$ is rigorous and we need only an upper bound.)

Step 3 — Bound $f$ and choose $\delta$

Combining $(2)$ with Step 2, for every $(x,y)\ne(0,0)$,

We want the right-hand side below $\varepsilon=0.01$. If $\sqrt{x^2+y^2}<\delta$, then $(3)$ gives $|f(x,y)-f(0,0)|<8\delta$. Demanding $8\delta\le 0.01$ forces

Take $\delta=0.00125\;\bigl(=\tfrac{1}{800}\bigr)$.

Step 4 — Verify

Let $0<\sqrt{x^2+y^2}<\delta=0.00125$. By $(3)$,

so the inequality $|f(x,y)-f(0,0)|<0.01$ holds; at $(x,y)=(0,0)$ it holds trivially since both sides are $0$. The required $\delta$ works.

(As a by-product, $(3)$ shows $f(x,y)\to 0=f(0,0)$ as $(x,y)\to(0,0)$, so the intended $f$ is continuous at the origin — which is exactly why a finite $\delta$ exists here, unlike the printed version flagged above.)

Answer