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UPSC 2026 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

Taylor's theorem with remainders · Calculus · asked 3× in 14 yrs · Read the full method →

Question

Using Maclaurin’s theorem, obtain the first four terms of the expansion of $e^{x\cos x}$ in ascending powers of $x$. Hence or otherwise find the limit of $\dfrac{e^x - e^{x\cos x}}{x - \sin x}$ as $x$ tends to zero.

Technique

Composition of standard Maclaurin series: substitute $t=x\cos x$ (itself expanded) into $e^{t}=\sum t^{k}/k!$ and collect powers of $x$. For the limit, expand numerator and denominator to the same order — here $x^3$ — and take the ratio of the leading coefficients. (L’Hôpital works but needs three differentiations of $e^{x\cos x}$; series is far cheaper and is what scores.)

Solution

Step 1 — Expand the exponent $t=x\cos x$.

Note $t=O(x)$, so to reach terms up to $x^{4}$ in $e^{t}$ we need $t,t^{2},t^{3},t^{4}$ only.

Step 2 — Powers of $t$ (retaining up to $x^{4}$).

Step 3 — Substitute into $e^{t}$.

Collect coefficients power by power:

• $x^{0}$: $\;1$
• $x^{1}$: $\;1$ (from $t$)
• $x^{2}$: $\;\dfrac12$ (from $t^{2}/2$)
• $x^{3}$: $\;-\dfrac12$ (from $t$) $+\dfrac16$ (from $t^{3}/6$) $=-\dfrac13$
• $x^{4}$: $\;-\dfrac12$ (from $t^{2}/2$) $+\dfrac1{24}$ (from $t^{4}/24$) $=-\dfrac{12}{24}+\dfrac{1}{24}=-\dfrac{11}{24}$

Step 4 — State the first four terms.

“First four terms in ascending powers of $x$” means the terms of degree $0,1,2,3$. All four of these are non-zero, so the alternative reading — “first four non-zero terms” — produces exactly the same four; there is no ambiguity to resolve here. The required answer is

i.e. the terms $1,\;x,\;\dfrac{x^{2}}{2},\;-\dfrac{x^{3}}{3}$. (The next term $-\tfrac{11}{24}x^{4}$ is not required; it was computed in Step 3 and is retained through Step 5 only to make the error term in the limit visible.)

Step 5 — The limit: expand numerator and denominator.

Denominator. $\sin x = x-\dfrac{x^{3}}{6}+\dfrac{x^{5}}{120}-\cdots$, so

The denominator vanishes to order 3; therefore the numerator must be expanded to $x^{3}$ at least — precisely the four terms already obtained. (This is why the first half of the question was asked.)

Numerator. $e^{x}=1+x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{6}+\dfrac{x^{4}}{24}+\cdots$, and from Step 3,

(the $1,\ x,\ \tfrac{x^{2}}{2}$ terms cancel exactly)

So the numerator also vanishes to order exactly $3$ (its leading coefficient $\tfrac12\neq0$) — the orders match and the limit is finite and non-zero.

Step 6 — Take the ratio.

Cancelling $x^{3}$ ($x\neq0$) and letting $x\to0$, both bracketed factors tend to $1$:

Numerical sanity check. The expansion predicts the ratio behaves like $3(1+x)$ near $0$; at $x=0.1$, numerator $\approx 5.520\times10^{-4}$ and denominator $=0.1-\sin(0.1)\approx1.6658\times10^{-4}$, giving $\approx 3.31 \approx 3(1+0.1)$ ✓ — the correct trend towards $3$.

Answer