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UPSC 2026 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Continuity of Functions on R; Epsilon-Delta · Real Analysis · asked 2× in 14 yrs · Read the full method →

Question

Find the points of discontinuity of the function $f : (0, 1) \to \mathbb{R}$ defined by

Technique

The expression is a function of the single quantity $t=1+\sin\frac{\pi}{x}$, and $\dfrac{t^{n}-1}{t^{n}+1}$ has three completely different limits according as $t>1$, $t=1$, $t<1$ — the classical ”$t^n$ trichotomy” that governs every “limit-of-$n$th-powers” function. Here $t\in[0,2]$, so the trichotomy is decided purely by the sign of $\sin\frac{\pi}{x}$, and the closed form collapses to $f=\operatorname{sgn}\!\left(\sin\frac{\pi}{x}\right)$. After that the question is elementary: locate the zeros of $\sin\frac\pi x$ in $(0,1)$ and check the two one-sided limits at each. The trigger: a limit of $n$th powers means “case-split on the base against $1$”, never a single algebraic manipulation.

Solution

Step 0 — Name the base and record its range; check the expression is even defined.

Put

Since $-1\le s\le 1$ for all $x$,

In particular $t\ge 0$, so $t^{n}\ge 0$ for every $n\ge1$, and therefore

The denominator never vanishes, for any $x\in(0,1)$ and any $n$: the sequence whose limit defines $f$ is well defined term by term. (This is worth one line — the base can be $0$, e.g. at $x=\tfrac23$ where $\sin\frac{3\pi}{2}=-1$, and a careless solution divides by something it has not checked.)

Step 1 — Evaluate the limit: the $t^{n}$ trichotomy.

Write $g_n(t)=\dfrac{t^{n}-1}{t^{n}+1}$.

Case (i): $t>1$, i.e. $s>0$. Then $t^{n}\to\infty$. Divide numerator and denominator by $t^{n}$ (legitimate, $t^n>0$):

Case (ii): $t=1$, i.e. $s=0$. Then $t^{n}=1$ for every $n$, so

Case (iii): $0\le t<1$, i.e. $-1\le s<0$. Then $t^{n}\to 0$ — for $0<t<1$ this is the standard geometric fact, and for $t=0$ (which does occur, at the points where $\sin\frac\pi x=-1$) we have $t^{n}=0$ for every $n\ge1$, so the limit is $0$ there too. Hence

Therefore

So $f$ takes only the three values $-1,0,1$; it is bounded, $|f|\le1$.

Step 2 — Locate the zeros of $\sin\frac{\pi}{x}$ inside $(0,1)$.

Substitute $u=\dfrac{\pi}{x}$. As $x$ runs over $(0,1)$, $u$ runs over $(\pi,\infty)$ — decreasingly in $x$: $x\downarrow$ corresponds to $u\uparrow$.

Combined with $u>\pi$ this forces $k\ge 2$, and

Hence

Note carefully that $x=1$ (which would come from $k=1$) is not in the open interval $(0,1)$, and neither is $x=0$; the open domain is doing real work here.

Step 3 — Sign of $\sin\frac{\pi}{x}$ between consecutive zeros; the explicit form of $f$.

For an integer $k\ge1$,

and on $(k\pi,(k+1)\pi)$ the sine has the constant sign $(-1)^{k}$ (positive on $(0,\pi)$ and alternating thereafter). Therefore

and $f(x)=0$ for $x\in E$. Since

this describes $f$ completely:

with $f\big(\tfrac1k\big)=0$ at every one of the separating points.

Step 4 — $f$ is continuous at every point of $(0,1)\setminus E$.

Let $x_{0}\in(0,1)$ with $x_{0}\notin E$. Then $x_{0}$ lies in exactly one of the open intervals $\left(\frac{1}{k+1},\frac1k\right)$, which is an open neighbourhood of $x_{0}$ inside $(0,1)$ on which $f$ is the constant $(-1)^{k}$. A function constant on a neighbourhood of a point is continuous at that point. Hence $f$ is continuous at $x_{0}$.

Step 5 — $f$ is discontinuous at every point of $E$; both one-sided limits computed.

Fix $k\ge2$ and put $x_{0}=\dfrac1k\in(0,1)$. By Step 3, $f(x_{0})=0$, and:

• Left-hand limit. For $x$ slightly less than $\frac1k$ we have $x\in\left(\frac{1}{k+1},\frac1k\right)$, where $f\equiv(-1)^{k}$. Hence

• Right-hand limit. For $x$ slightly greater than $\frac1k$ we have $x\in\left(\frac1k,\frac{1}{k-1}\right)$ (this interval is non-empty and contained in $(0,1)$ precisely because $k\ge2$), where $f\equiv(-1)^{k-1}$. Hence

The two one-sided limits are $+1$ and $-1$ in one order or the other; in particular

so $\displaystyle\lim_{x\to 1/k}f(x)$ does not exist, and $f$ is discontinuous at $x_{0}=\frac1k$.

Both one-sided limits exist and are finite, so each is a discontinuity of the first kind (a jump), with jump

and the value $f(x_{0})=0$ is the midpoint of the jump — it agrees with neither one-sided limit, so the discontinuity is irremovable no matter how $f(1/k)$ is redefined.

Concretely, at $x_{0}=\tfrac12$ ($k=2$): $\lim_{x\to\frac12^{-}}f=+1$, $\ f\left(\tfrac12\right)=0$, $\ \lim_{x\to\frac12^{+}}f=-1$.

Step 6 — Assemble, and note what the open interval excludes.

Two structural remarks that the phrasing of the question invites:

1. $D$ is countably infinite, and its points accumulate only at $x=0$, which is not in the domain. Consequently every point of $D$ is isolated in $D$, and on any closed subinterval $[a,b]\subset(0,1)$ with $a>0$ the function $f$ has only finitely many discontinuities — so $f$ is Riemann integrable on every such $[a,b]$ (Lebesgue’s criterion: $D$ has measure zero).
2. Had the domain been $[0,1)$, the point $x=0$ would be a further, far worse discontinuity: $f$ oscillates between $+1$ and $-1$ in every neighbourhood of $0$, so neither one-sided limit exists there (a discontinuity of the second kind). Had it been $(0,1]$, the endpoint $x=1$ would join $D$ as well, since $\sin\pi=0$ gives $f(1)=0$ while $f\to-1$ as $x\to1^{-}$. The open interval $(0,1)$ excludes both. $\qquad\blacksquare$

Answer