UPSC 2024 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Continuity of real functions · Calculus · asked 2× in 14 yrs · Read the full method →

Question

Discuss the continuity of the function

f(x)=\begin{cases}\dfrac{1}{1-e^{-1/x}}, & x\ne 0\\ 0, & x=0\end{cases}

for all values of $x$ .

Technique

One-sided limits at $x=0$ by examining the sign of $-1/x$ as $x\to 0^\pm$ .

Solution

$Graph of f(x)=1/(1-e^{-1/x}) near x=0. The left branch (x<0) rises from 0 toward the origin, where f(0)=0 is marked by a filled dot; the right branch (x>0) descends toward height 1, marked by an open circle at (0,1) since the value 1 is approached but not attained from the right. The two one-sided limits differ (0 on the left, 1 on the right), exhibiting a jump discontinuity at x=0.$

Step 1 — Continuity for $x\ne 0$ .

For $x\ne 0$ , the exponent $-1/x$ is finite and non-zero, so $e^{-1/x}\ne 1$ , making the denominator $1-e^{-1/x}\ne 0$ . The function is a composition of elementary continuous functions, hence continuous at every $x\ne 0$ .

Step 2 — One-sided limits at $x=0$ .

As $x\to 0^+$ : $-1/x\to-\infty$ , so $e^{-1/x}\to 0$ , giving $f(x)\to \dfrac{1}{1-0}=1$ .

As $x\to 0^-$ : $-1/x\to+\infty$ , so $e^{-1/x}\to+\infty$ , giving $f(x)\to \dfrac{1}{1-\infty}=0$ .

Step 3 — Compare with $f(0)=0$ .

$\lim_{x\to 0^-}f(x)=0=f(0)$ : $f$ is left-continuous at $0$ .
$\lim_{x\to 0^+}f(x)=1\ne 0=f(0)$ : $f$ is not right-continuous at $0$ .

The two one-sided limits exist but are unequal, so the two-sided limit does not exist.

Answer

\boxed{\;f\text{ is continuous for all }x\ne 0;\;f\text{ has a jump discontinuity at }x=0\text{ (left-continuous but not right-continuous).}\;}