Limit JEE 2024: Solution Using Limits and Quadratics

What is the answer to the limit JEE 2024 question?

The correct numerical entry for this limit JEE 2024 question is 6, not 0: both limits equal 1, but the fixed constant term determines the quadratic’s scale. The supplied question bank labels this a JEE Main 2024 numerical-answer question, tagged Hard.

The quadratic uses a different variable from the limits to avoid confusion. Define

α = {lim}_{x \to 0^{+}} \frac{\sqrt{x \tan x} - e^{x}}{\sqrt{x \tan x} - (\sqrt{2})^{x}}, β = {lim}_{x \to 0} (1 + \sin x)^{1 / \cot x} .

These are the roots of $a t^{2} + b t - \sqrt{e} = 0 .$

Determine $12 \ln (a + b) .$

Follow the official solution’s order: evaluate beta using logarithms, evaluate alpha by direct limits, then match coefficients. Finding the roots is not enough: the stated constant term fixes the coefficients needed for the requested logarithm.

How do you evaluate beta by taking logarithms?

Beta equals 1 because the logarithm of the variable-power expression tends to zero. The exponent itself tends to zero, not infinity, so this is not the standard indeterminate power form $1^{\infty} .$

For sufficiently small nonzero inputs, rewrite the reciprocal cotangent and set

\frac{1}{\cot x} = \tan x, y = (1 + \sin x)^{\tan x} .

The base is positive near zero, so the real logarithm is defined. Taking logarithms brings the exponent down: $\ln y = \tan x \cdot \ln (1 + \sin x) .$

Apply the standard equivalents:

\tan x ~ x, \sin x ~ x, \ln (1 + u) ~ u (u \to 0) .

Together, they give $\ln y ~ x \cdot \sin x ~ x^{2} ⟶ 0 .$

Return to the original expression using continuity of the exponential function:

y = \exp (\ln y) ⟶ \exp (0) = 1, β = 1 .

Inspect the actual exponent before choosing a familiar limit formula. A base approaching 1 alone does not determine the limiting form; here, the logarithm and the exponent both approach zero.

How do you evaluate alpha without differentiating?

Alpha also equals 1, since its numerator and denominator both tend to negative one. The denominator’s limit is nonzero, so the quotient limit law applies directly. Check this limiting form before attempting differentiation.

Use the tangent equivalent to handle the square root:

\tan x ~ x ⟹ \sqrt{x \tan x} ~ \sqrt{x^{2}} = | x | .

Because the approach is from the right,

x \to 0^{+} ⟹ | x | = x, \sqrt{x \tan x} \to 0 .

The exponential terms satisfy

e^{x} \to 1, (\sqrt{2})^{x} = 2^{x / 2} \to 1 .

Therefore, the numerator and denominator limits are

{lim}_{x \to 0^{+}} (\sqrt{x \tan x} - e^{x}) = - 1,

{lim}_{x \to 0^{+}} (\sqrt{x \tan x} - (\sqrt{2})^{x}) = - 1 .

Hence

α = \frac{- 1}{- 1} = 1 .

Neither L’Hôpital’s rule nor a higher-order expansion is needed. L’Hôpital’s rule does not apply to this nonzero-over-nonzero limiting form.

How does the constant term fix the quadratic’s coefficients?

Matching the constant term fixes the quadratic’s scale. The roots determine the polynomial only up to a nonzero multiplier, but the question specifies that multiplier through the constant coefficient. Keep the original leading coefficient when writing the repeated-root form.

Since both roots are 1,

α = β = 1, a t^{2} + b t - \sqrt{e} = a (t - 1)^{2} .

Expand without changing the leading coefficient:

a (t - 1)^{2} = a t^{2} - 2 a t + a .

Compare the constant coefficients first, then the linear coefficients:

a = - \sqrt{e}, b = - 2 a = 2 \sqrt{e} .

The leading coefficient is negative, while the linear coefficient is positive. Their sum is

a + b = - \sqrt{e} + 2 \sqrt{e} = \sqrt{e} > 0 .

This confirms that the requested real logarithm is defined. Complete the numerical calculation:

12 \ln (a + b) = 12 \ln (\sqrt{e}) = 12 \times \frac{1}{2} = 6 .

As a substitution check, evaluate the recovered polynomial at its repeated root:

t = 1 : - \sqrt{e} + 2 \sqrt{e} - \sqrt{e} = 0 .

The coefficient comparison fixes the scale; the substitution check confirms the root. Verifying a root alone cannot distinguish different nonzero multiples of the same polynomial.

Why can normalising the polynomial give the wrong answer 0?

Zero can arise when normalised coefficients are reused as the original coefficients. This is a numerical-answer question: 0 is an incorrect result to diagnose, not a supplied multiple-choice option.

The invalid shortcut is to replace the required polynomial with

- (t - 1)^{2} = - t^{2} + 2 t - 1 .

Then incorrectly read off

a = - 1, b = 2, a + b = 1, 12 \ln 1 = 0 .

This replacement changes the stipulated constant coefficient from

- \sqrt{e} to - 1 .

Multiplying a polynomial by a nonzero constant preserves its roots, but changes its coefficients and their sum. Dividing the original equation by $\sqrt{e}$ is legitimate when finding roots. It is not legitimate to treat the resulting coefficients as the original ones when evaluating the requested expression.

Same roots does not mean same coefficient sum. Before using coefficients in a separate expression, check them against the original polynomial, not its normalised version.

Which three related limit questions should you try next?

These original related practice questions test the exponent, the one-sided square root and the coefficient scale separately. They belong to Limit, Continuity and Differentiability, but are not authenticated PYQs; no additional past-paper questions are supplied here. Attempt all three prompts before reading the checks.

Question 1: Evaluate

{lim}_{x \to 0} (1 + \sin x)^{1 / x} .

Question 2: Evaluate

{lim}_{x \to 0^{-}} \frac{\sqrt{x \tan x}}{x} .

Question 3: Let the following limit be a repeated root of the given quadratic. Find the requested logarithm.

r = {lim}_{x \to 0} \frac{e^{x} - 1}{x}, A t^{2} + B t - e^{2} = 0, find \ln (A + B) .

Worked checks

Question 1: Set the expression equal to a positive variable and take logarithms:

y = (1 + \sin x)^{1 / x}, \ln y = \frac{\ln (1 + \sin x)}{x} = \frac{\ln (1 + \sin x)}{\sin x} \frac{\sin x}{x} ⟶ 1 .

y \to e .

Here the reciprocal exponent is unbounded in magnitude, unlike the original tangent exponent, which tends to zero.

Question 2: Retain the absolute value until applying the left-hand condition:

\frac{\sqrt{x \tan x}}{x} = \frac{| x |}{x} \sqrt{\frac{\tan x}{x}} ⟶ (- 1) (1) = - 1 .

The original approach was right-handed; this one changes the sign.

Question 3: The standard exponential limit gives the repeated root. Match the constant term:

r = 1, A (t - 1)^{2} = A t^{2} + B t - e^{2} .

A = - e^{2}, B = 2 e^{2}, \ln (A + B) = \ln (e^{2}) = 2 .

Before accepting any coefficient-based answer, expand your recovered polynomial and check its constant term against the question.

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

What is the answer to the limit JEE 2024 question?

The correct numerical answer is 6. Both limits equal 1, and matching the quadratic’s constant term gives a = -√e and b = 2√e. Therefore, 12 ln(a+b) = 12 ln(√e) = 6.

How do I evaluate beta in the JEE 2024 limit question?

Rewrite 1/cot x as tan x and take logarithms of y = (1 + sin x)^(tan x). Then ln y = tan x · ln(1 + sin x), which tends to zero as x tends to zero. Hence beta = 1; the exponent tends to zero, so this is not a 1-to-infinity indeterminate form.

Can I use L’Hôpital’s rule for alpha in this question?

No: alpha’s numerator and denominator both tend to -1, so the expression is not an indeterminate form to which L’Hôpital’s rule applies. The quotient limit law gives alpha = 1 directly.

How do I find a and b when both roots are 1?

Write the original polynomial as at² + bt - √e = a(t - 1)². Expanding and matching constant terms gives a = -√e, while matching linear coefficients gives b = -2a = 2√e.

Why is 0 the wrong answer to this limit question?

The answer 0 comes from replacing the original polynomial with -(t - 1)² and incorrectly using its coefficients as a and b. That replacement changes the required constant term from -√e to -1. Scaling preserves roots, but it does not preserve the coefficient sum used in 12 ln(a+b).

Limit JEE 2024: Why the Answer Is 6, Not 0

What is the answer to the limit JEE 2024 question?

How do you evaluate beta by taking logarithms?

How do you evaluate alpha without differentiating?

How does the constant term fix the quadratic’s coefficients?

Why can normalising the polynomial give the wrong answer 0?

Which three related limit questions should you try next?

Frequently asked questions

What is the answer to the limit JEE 2024 question?

How do I evaluate beta in the JEE 2024 limit question?

Can I use L’Hôpital’s rule for alpha in this question?

How do I find a and b when both roots are 1?

Why is 0 the wrong answer to this limit question?

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