Differentiation Rules JEE 2025: Worked Tangent-Sum Limit

What is the answer to the infinite tangent-sum limit?

The supplied solution selects option D for this worked differentiation rules jee 2025 problem, with limiting value

- 2 .

The decisive step is finding the denominator’s leading coefficient using the supplied Taylor-series method.

Source note: The bank labels the year 2025 but records 2023-01-24, Morning. The supplied material does not resolve this discrepancy, so this is not a securely dated 2025 paper question.

Define the summand and function using separate symbols for the running index and upper limit:

u_{k} (x) = \frac{\tan (x / 2^{k}) - \tan (x / 2^{k + 1})}{1 - \tan (x / 2^{k}) \tan (x / 2^{k + 1})}, f (x) = {lim}_{N \to \infty} \sum_{k = 1}^{N} u_{k} (x) .

This repairs the source’s repeated index without changing its meaning. Find the value approached by

\frac{e^{- x} - 1}{x - f (x)} as x \to 0 .

The full option texts are absent; only the supplied value and answer letter can be checked.

How do you expand one summand without missing the denominator?

Expand the numerator and the reciprocal denominator separately, then multiply. The reciprocal denominator contributes to the cubic coefficient, so replacing it with one loses a term in the supplied calculation. Taylor expansion also avoids differentiating the infinite sum without justification.

Set

a = \frac{x}{2^{k}}, \frac{x}{2^{k + 1}} = \frac{a}{2} .

The tangent expansions are

\tan a = a + \frac{a^{3}}{3} + O (a^{5}), \tan \frac{a}{2} = \frac{a}{2} + \frac{a^{3}}{24} + O (a^{5}) .

Subtracting gives the numerator:

\tan a - \tan \frac{a}{2} = \frac{a}{2} + (\frac{1}{3} - \frac{1}{24}) a^{3} + O (a^{5}) = \frac{a}{2} + \frac{7 a^{3}}{24} + O (a^{5}) .

For the denominator, multiply first:

\tan a \tan \frac{a}{2} = (a + \frac{a^{3}}{3} + O (a^{5})) (\frac{a}{2} + \frac{a^{3}}{24} + O (a^{5})) = \frac{a^{2}}{2} + O (a^{4}) .

Hence

1 - \tan a \tan \frac{a}{2} = 1 - \frac{a^{2}}{2} + O (a^{4}) .

Use the reciprocal expansion:

\frac{1}{1 - z} = 1 + z + O (z^{2}), \frac{1}{1 - a^{2} / 2 + O (a^{4})} = 1 + \frac{a^{2}}{2} + O (a^{4}) .

The reciprocal’s quadratic term is positive. Multiplication now gives

u_{k} = (\frac{a}{2} + \frac{7 a^{3}}{24} + O (a^{5})) (1 + \frac{a^{2}}{2} + O (a^{4})) = \frac{a}{2} + (\frac{7}{24} + \frac{1}{4}) a^{3} + O (a^{5}) = \frac{a}{2} + \frac{13 a^{3}}{24} + O (a^{5}) .

The fifth-order remainder notation means that, near zero, its magnitude is bounded by a constant times the absolute fifth power: $| R (a) | \leq C | a |^{5} .$

How do you sum the expansion and justify the remainder?

The linear terms sum to half the input, and the cubic terms form a separate geometric series. The remainder is summable too: one constant bounds it for every index. This bound justifies passing from the local Taylor expansion to the infinite sum.

Substitution gives

u_{k} (x) = \frac{x}{2^{k + 1}} + \frac{13 x^{3}}{24} 8^{- k} + O (| x |^{5} 32^{- k}) .

Because the index starts at one, the linear terms begin as

\frac{x}{4} + \frac{x}{8} + \frac{x}{16} + \dots, not \frac{x}{2} + \frac{x}{4} + \dots .

Evaluate both geometric sums with their actual first terms:

\sum_{k = 1}^{\infty} 2^{- k - 1} = \frac{1 / 4}{1 - 1 / 2} = \frac{1}{2}, \sum_{k = 1}^{\infty} 8^{- k} = \frac{1 / 8}{1 - 1 / 8} = \frac{1}{7} .

For sufficiently small input, every scaled argument lies in one fixed neighbourhood of zero. The Taylor remainder therefore admits a constant independent of the index:

| R_{k} (x) | \leq C | x |^{5} 32^{- k} .

Since

\sum_{k = 1}^{\infty} 32^{- k} = \frac{1 / 32}{1 - 1 / 32} = \frac{1}{31}, \sum_{k = 1}^{\infty} | R_{k} (x) | \leq \frac{C}{31} | x |^{5},

the total remainder remains fifth order. Summing the finite partial sums and then taking their limit gives

f (x) = \frac{x}{2} + \frac{13 x^{3}}{168} + O (x^{5}) .

How does normalising the quotient give option D?

Divide numerator and denominator by the input. The normalised numerator tends to negative one, while the normalised denominator tends to positive one-half. Their ratio gives the supplied answer without L’Hôpital’s rule or termwise differentiation.

For nonzero input,

\frac{f (x)}{x} = \frac{1}{2} + \frac{13 x^{2}}{168} + O (x^{4}) ⟶ \frac{1}{2} .

Rewrite the quotient exactly:

\frac{e^{- x} - 1}{x - f (x)} = \frac{(e^{- x} - 1) / x}{1 - f (x) / x} .

Using the standard exponential limit,

\frac{e^{- x} - 1}{x} ⟶ - 1, 1 - \frac{f (x)}{x} ⟶ 1 - \frac{1}{2} = \frac{1}{2} .

Therefore,

{lim}_{x \to 0} \frac{e^{- x} - 1}{x - f (x)} = \frac{- 1}{1 / 2} = - 2 .

This verifies option D according to the supplied solution. Only the linear coefficient controls this limit; retaining the cubic term follows the official calculation and supports the cancellation exercise below.

Why is assigning negative one to the denominator limit wrong?

The normalised denominator tends to positive one-half, not negative one. A minus sign between two expressions does not determine their difference’s leading coefficient. The supplied solution associates the previously keyed option A with this incorrect assignment:

\frac{x - f (x)}{x} ⟶ - 1 .

That would produce

\frac{- 1}{- 1} = 1,

but this diagnoses the stated method error, not unseen option text. The demonstrated calculation instead gives

\frac{x - f (x)}{x} = 1 - \frac{f (x)}{x} ⟶ \frac{1}{2} .

Both linear contributions must be retained. Direct subtraction confirms it:

x - f (x) = \frac{x}{2} - \frac{13 x^{3}}{168} + O (x^{5}) .

Thus the normalised denominator is positive near zero, even though the unnormalised expression changes sign with the input.

The tangent subtraction identity does not apply here:

\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} .

Its denominator has a plus sign, unlike the given summand’s minus sign.

How do you solve related slope, cancellation and numerator questions?

Use the same expansion, but check what survives before choosing its required order. These are original related practice questions, not verified PYQs, from Limit, Continuity and Differentiability. Attempt each before reading its worked check.

What is the derivative at zero?

Question 1: For the same function, with its value at zero set to zero, the derivative is one-half. Check this directly from the derivative definition:

f (0) = 0, f^{'} (0) = {lim}_{x \to 0} \frac{f (x) - f (0)}{x} = {lim}_{x \to 0} (\frac{1}{2} + \frac{13 x^{2}}{168} + O (x^{4})) = \frac{1}{2} .

The definition extracts the linear coefficient. No differentiation of the infinite sum is required.

What happens when the linear term is cancelled?

Question 2: Subtracting the linear term makes the cubic coefficient determine the following limit:

{lim}_{x \to 0} \frac{f (x) - x / 2}{x^{3}} = {lim}_{x \to 0} (\frac{13}{168} + O (x^{2})) = \frac{13}{168} .

The cubic term is now essential. A first-order expansion alone would not determine this answer.

What if the exponential contains a fixed real constant?

Question 3: The normalised numerator tends to the negative of the constant, while the denominator still tends to one-half. For a fixed real constant, evaluate

{lim}_{x \to 0} \frac{e^{- b x} - 1}{x - f (x)}, b \in R .

The worked check is

\frac{e^{- b x} - 1}{x} ⟶ - b, {lim}_{x \to 0} \frac{e^{- b x} - 1}{x - f (x)} = \frac{- b}{1 / 2} = - 2 b .

When the constant is zero, the numerator is identically zero and the answer is zero. Before deciding how far to expand, determine the first nonzero term after any cancellation.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

How do you solve the infinite tangent-sum limit without L'Hôpital's rule?

Taylor expansion and geometric summation give the tangent-sum function as f(x) = x/2 + 13x³/168 + O(x⁵). Divide both parts of (e^(-x) - 1)/(x - f(x)) by x: the numerator tends to -1 and the denominator to 1/2. The limit is therefore -2, matching option D in the supplied solution.

Why can't I use the tan(A - B) identity in this problem?

The tangent subtraction identity has denominator 1 + tan(A)tan(B). The given summand has 1 - tan(A)tan(B), so the identity does not apply. Expand the numerator and reciprocal denominator separately instead.

Do I need the cubic term to find this limit?

No: the original limit depends only on the linear coefficient in f(x) = x/2 + 13x³/168 + O(x⁵). The cubic term becomes essential when the linear term is cancelled. For example, (f(x) - x/2)/x³ tends to 13/168.

Is this tangent-sum question a verified JEE 2025 PYQ?

The source bank labels the year as 2025 but records the session as 2023-01-24, Morning. The supplied material does not resolve this discrepancy, so the question cannot be securely identified as a 2025 PYQ. The three related practice questions are original exercises, not verified PYQs.