← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

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

Question

Let $f:D(\subseteq\mathbb R^2)\to\mathbb R$ be a function and $(a,b)\in D$. If $f(x,y)$ is continuous at $(a,b)$, then show that the functions $f(x,b)$ and $f(a,y)$ are continuous at $x=a$ and at $y=b$ respectively.

Technique

$\varepsilon$–$\delta$ definition of continuity in $\mathbb R^2$; restrict the joint $\delta$ to a horizontal/vertical line where distance reduces to a single-coordinate modulus.

Solution

Step 1 — State the hypothesis precisely

$f$ is continuous at $(a,b)$ means: for every $\varepsilon>0$ there exists $\delta>0$ such that for all $(x,y)\in D$,

Define the section functions $g(x):=f(x,b)$ (fixing $y=b$) and $h(y):=f(a,y)$ (fixing $x=a$), on the slices where they are defined.

Step 2 — Continuity of $g(x)=f(x,b)$ at $x=a$

Fix $\varepsilon>0$ and take the $\delta>0$ from $(\ast)$. Suppose $|x-a|<\delta$. Consider the point $(x,b)$: its distance to $(a,b)$ is

So $(x,b)$ satisfies the hypothesis of $(\ast)$, giving

Thus for every $\varepsilon>0$ there is $\delta>0$ with $|x-a|<\delta\Rightarrow|g(x)-g(a)|<\varepsilon$, i.e. $g$ is continuous at $x=a$.

Step 3 — Continuity of $h(y)=f(a,y)$ at $y=b$

Symmetrically, fix $\varepsilon>0$ and the same $\delta$. If $|y-b|<\delta$, the point $(a,y)$ has distance $\sqrt{0+(y-b)^2}=|y-b|<\delta$ from $(a,b)$, so by $(\ast)$

Hence $h$ is continuous at $y=b$.

Answer