JEE Main 2026 (22 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :

Q: Wait, why can't we just use L'Hopital's rule? It's a $0/0$ form, right?

You are absolutely correct that it is a $0/0$ form! However, L'Hopital's rule requires differentiating the numerator and denominator multiple times. Given the products and compositions of functions like $e^{(a-1)x}$ and $\log_e(1+x)$, the derivatives will become increasingly complex and prone to algebraic errors. Taylor series allows us to replace these functions with simple polynomials, making the limit evaluation much cleaner and more robust.

Q: Why do we set the constant and $x$ coefficients to zero in the numerator?

If the numerator had a non-zero constant term or an $x$ term, the limit would be of the form $\frac{\text{constant}}{x^2}$ or $\frac{x}{x^2}$, which would approach infinity as $x \to 0$. Since we are given that the limit is a finite value (2), the numerator must have the same leading power of $x$ as the denominator. Forcing those lower-degree coefficients to zero is the only way to ensure the limit converges to a finite value.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Look at the denominator

x cos x - lo g_{e} (1 + x)

. As

x \to 0

, both terms approach zero. Instead of using L'Hopital's rule, which will lead to a nightmare of product and chain rules, visualize the Taylor series expansion for each function around

x = 0

. This is the geometric key to unlocking the problem.

Hint 2

Hint 2 (The Arsenal): Your toolkit must include the standard Taylor series expansions:

cos x = 1 - \frac{x ^{2}}{2 !} + \dots

lo g_{e} (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots

, and

e^{t} = 1 + t + \frac{t ^{2}}{2 !} + \dots

. Apply these to both the numerator and the denominator to express the limit as a ratio of polynomials in

x

Hint 3

Hint 3 (The Execution): Expand the denominator to find its leading term. You will find that the

x

terms cancel out, leaving

\frac{x ^{2}}{2}

as the dominant term. For the limit to be a finite non-zero value (which is 2), the numerator must also have

x^{2}

as its lowest power. Therefore, you must set the coefficients of the constant term and the

x

term in the numerator to zero.

Visualized Solution

Understanding the Limit Form

Given limit: $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$
Objective: Find the value of $a^{2} + b^{2} + c^{2}$
Notice the form is $\frac{0}{0}$ as $x \to 0$ .

The Taylor Series Strategy

To evaluate this limit, we use Taylor series expansion around $x = 0$ .
We will expand both numerator and denominator up to the required power of $x$ .

Expanding the Denominator

Denominator: $x cos x - lo g_{e} (1 + x)$
$cos x = 1 - \frac{x ^{2}}{2 !} + \dots ⟹ x cos x = x - \frac{x ^{3}}{2} + \dots$
$lo g_{e} (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$

Simplifying the Denominator

Subtracting the two expansions:
$(x - \frac{x ^{3}}{2}) - (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3})$
The $x$ terms cancel out.
Leading term becomes $\frac{x ^{2}}{2}$ .

Expanding the Numerator: $e^{(a - 1) x}$

First term: $e^{(a - 1) x}$
Using $e^{t} = 1 + t + \frac{t ^{2}}{2 !} + \dots$
Substitute $t = (a - 1) x$ :
$1 + (a - 1) x + \frac{( a - 1 ) ^{2} x ^{2}}{2} + \dots$

Expanding the Numerator: $2 cos b x$

Second term: $2 cos b x$
Using $cos t = 1 - \frac{t ^{2}}{2 !} + \dots$
Substitute $t = b x$ :
$2 (1 - \frac{b ^{2} x ^{2}}{2}) = 2 - b^{2} x^{2} + \dots$

Expanding the Numerator: $(c - 2) e^{- x}$

Third term: $(c - 2) e^{- x}$
Using $e^{- x} = 1 - x + \frac{x ^{2}}{2 !} - \dots$
Multiply by $(c - 2)$ :
$(c - 2) - (c - 2) x + \frac{( c - 2 ) x ^{2}}{2} - \dots$

Grouping Numerator Terms

Combine all expanded terms and group by powers of $x$ .
Constant term: $1 + 2 + (c - 2) = c + 1$
Coefficient of $x$ : $(a - 1) - (c - 2) = a - c + 1$
Coefficient of $x^{2}$ : $\frac{( a - 1 ) ^{2}}{2} - b^{2} + \frac{c - 2}{2}$

Condition for a Finite Limit

The denominator's leading term is $\frac{x ^{2}}{2}$ .
For the limit to exist and equal $2$ , the numerator must also have $x^{2}$ as its lowest power.
Therefore, the constant term and the coefficient of $x$ must be zero.

Solving for $c$

Equate the constant term to zero:
$c + 1 = 0$
$⟹ c = - 1$

Solving for $a$

Equate the coefficient of $x$ to zero:
$a - c + 1 = 0$
Substitute $c = - 1$ : $a - (- 1) + 1 = 0$
$⟹ a + 2 = 0 ⟹ a = - 2$

Setting up the Limit Equation

The limit is the ratio of the coefficients of $x^{2}$ .
$\frac{Coefficient of x ^{2} in Numerator}{Coefficient of x ^{2} in Denominator} = 2$
$\frac{\frac{( a - 1 ) ^{2}}{2} - b ^{2} + \frac{c - 2}{2}}{1/2} = 2$
$⟹ \frac{( a - 1 ) ^{2}}{2} - b^{2} + \frac{c - 2}{2} = 1$

Solving for $b^{2}$

Substitute $a = - 2$ and $c = - 1$ :
$\frac{( - 2 - 1 ) ^{2}}{2} - b^{2} + \frac{- 1 - 2}{2} = 1$
$\frac{9}{2} - b^{2} - \frac{3}{2} = 1$
$3 - b^{2} = 1 ⟹ b^{2} = 2$

Final Calculation

We need to find $a^{2} + b^{2} + c^{2}$ .
$a^{2} = (- 2)^{2} = 4$
$b^{2} = 2$
$c^{2} = (- 1)^{2} = 1$
Sum $= 4 + 2 + 1 = 7$

00:00 / 00:00

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Solution Diagram

Analyzing the Setup

Imagine you are standing on the edge of a mathematical cliff. You are staring at a limit:

x \to 0 lim \frac{e ^{(a - 1) x} + 2 cos b x + ( c - 2 ) e ^{- x}}{x cos x - lo g _{e} ( 1 + x )} = 2

It looks intimidating, but in the world of JEE Advanced, complexity is often just a mask for elegance. Let's peel it back.

The Denominator's Secret

First, let's look at the denominator:

x cos x - lo g_{e} (1 + x)

. If we plug in

x = 0

, we get

0 - 0 = 0

, which is a

0/0

indeterminate form.

Instead of rushing into the brute force of L'Hopital's rule, let's use the scalpel of Taylor series. We know that:

cos x = 1 - \frac{x ^{2}}{2 !} + \dots ⟹ x cos x = x - \frac{x ^{3}}{2} + \dots

lo g_{e} (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots

When we subtract these, the

x

terms vanish. We are left with:

(x - \frac{x ^{3}}{2}) - (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3}) = \frac{x ^{2}}{2} - \frac{5 x ^{3}}{6} + \dots

The leading term is

\frac{x ^{2}}{2}

. This is our anchor.

Taming the Numerator

Now, we must ensure the numerator behaves. We expand each term using Taylor series:

e^{(a - 1) x} = 1 + (a - 1) x + \frac{( a - 1 ) ^{2} x ^{2}}{2} + \dots

2 cos b x = 2 (1 - \frac{b ^{2} x ^{2}}{2} + \dots) = 2 - b^{2} x^{2} + \dots

(c - 2) e^{- x} = (c - 2) (1 - x + \frac{x ^{2}}{2} + \dots) = (c - 2) - (c - 2) x + \frac{( c - 2 ) x ^{2}}{2} + \dots

Now, we group them by powers of

x

. The constant term is

1 + 2 + (c - 2) = c + 1

. The coefficient of

x

(a - 1) - (c - 2) = a - c + 1

. The coefficient of

x^{2}

is:

\frac{( a - 1 ) ^{2}}{2} - b^{2} + \frac{c - 2}{2}

The Constraint of Finiteness

For the limit to be

2

, the numerator cannot have any constant or

x

terms. If it did, the limit would blow up to infinity. We force them to zero:

c + 1 = 0 ⟹ c = - 1

a - c + 1 = 0 ⟹ a - (- 1) + 1 = 0 ⟹ a = - 2

With these values, the limit becomes the ratio of the

x^{2}

coefficients divided by the leading coefficient of the denominator:

\frac{\frac{( a - 1 ) ^{2}}{2} - b ^{2} + \frac{c - 2}{2}}{1/2} = 2

Substituting

a = - 2

and

c = - 1

, we get:

\frac{\frac{( - 3 ) ^{2}}{2} - b ^{2} + \frac{- 3}{2}}{1/2} = 2 ⟹ \frac{9}{2} - b^{2} - \frac{3}{2} = 1 ⟹ 3 - b^{2} = 1 ⟹ b^{2} = 2

Final Calculation

We have found

a = - 2

b^{2} = 2

, and

c = - 1

. The question asks for

a^{2} + b^{2} + c^{2}

(- 2)^{2} + 2 + (- 1)^{2} = 4 + 2 + 1 = 7

The final answer is 7.

Common Traps & FAQs

Wait, why can't we just use L'Hopital's rule? It's a

0/0

form, right?

You are absolutely correct that it is a

0/0

form! However, L'Hopital's rule requires differentiating the numerator and denominator multiple times. Given the products and compositions of functions like

e^{(a - 1) x}

and

lo g_{e} (1 + x)

, the derivatives will become increasingly complex and prone to algebraic errors. Taylor series allows us to replace these functions with simple polynomials, making the limit evaluation much cleaner and more robust.

How do I know how many terms to expand in the Taylor series?

Great question! The rule of thumb is to expand until you reach the first non-zero term in the denominator. In this problem, the denominator's leading term is

\frac{x ^{2}}{2}

. This tells us that we need to expand the numerator up to the

x^{2}

term. Any higher-order terms (like

x^{3}, x^{4}

) will vanish as

x \to 0

, so we can safely ignore them.

Why do we set the constant and

x

coefficients to zero in the numerator?

If the numerator had a non-zero constant term or an

x

term, the limit would be of the form

\frac{constant}{x ^{2}}

\frac{x}{x ^{2}}

, which would approach infinity as

x \to 0

. Since we are given that the limit is a finite value (2), the numerator must have the same leading power of

x

as the denominator. Forcing those lower-degree coefficients to zero is the only way to ensure the limit converges to a finite value.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :

Select Answer:

Visualized Solution

Understanding the Limit Form

The Taylor Series Strategy

Expanding the Denominator

Simplifying the Denominator

Expanding the Numerator: $e^{(a - 1) x}$

Expanding the Numerator: $2 cos b x$

Expanding the Numerator: $(c - 2) e^{- x}$

Grouping Numerator Terms

Condition for a Finite Limit

Solving for $c$

Solving for $a$

Setting up the Limit Equation

Solving for $b^{2}$

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Denominator's Secret

Taming the Numerator

The Constraint of Finiteness

Final Calculation

Similar Questions

$lim_{x \to 0} \frac{a x ^{2} e ^{x} - b l o g _{e} ( 1 + x ) + c x e ^{- x}}{x ^{2} s i n x} = 1$ then $16 (a^{2} + b^{2} + c^{2})$ is equal to _______.

If $lim_{x \to 0} \frac{a e ^{x} - b c o s x + c e ^{- x}}{x s i n x} = 2$ , then $a + b + c$ is equal to ______

If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then $2 α - β$ is equal to :

If $lim_{x \to 0} \frac{e ^{a x} - c o s ( b x ) - \frac{c x e ^{- c x}}{2}}{1 - c o s ( 2 x )} = 17$ , then $5 a^{2} + b^{2}$ is equal to

For $α, β, γ, \in R$ , if $lim_{x \to 0} \frac{x ^{2} s i n α x + ( γ - 1 ) e ^{x^{2}}}{s i n 2 x - β x} = 3$ , then $β + γ - α$ is equal to:

If $lim_{x \to 0} \frac{α x e ^{x} - β l o g _{e} ( 1 + x ) + γ x ^{2} e ^{- x}}{x s i n ^{2} x} = 10$ , $α, β, γ \in R$ , then the value of $α + β + γ$ is

If the value of $lim_{x \to 0} (2 - cos x cos 2 x)^{\frac{x + 2}{x ^{2}}}$ is equal to $e^{a}$ , then $a$ is equal to

If $lim_{x \to 0} \frac{α e ^{x} + β e ^{- x} + γ s i n x}{x s i n ^{2} x} = \frac{2}{3}$ , where $α, β, γ \in R$ , then which of the following is NOT correct ?

$lim_{x \to 0} (\frac{3 x ^{2} + 2}{7 x ^{2} + 2})^{\frac{1}{x ^{2}}}$ is equal to

If $lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}$ , then the values of $a$ and $b$ , are

Select Answer:

Visualized Solution

Understanding the Limit Form

The Taylor Series Strategy

Expanding the Denominator

Simplifying the Denominator

Expanding the Numerator: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e(a−1)x

Expanding the Numerator: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2cosbx

Expanding the Numerator: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (c−2)e−x

Grouping Numerator Terms

Condition for a Finite Limit

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Setting up the Limit Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Denominator's Secret

Taming the Numerator

The Constraint of Finiteness

Final Calculation

Similar Questions

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Expanding the Numerator: $e^{(a - 1) x}$

Expanding the Numerator: $2 cos b x$

Expanding the Numerator: $(c - 2) e^{- x}$

Solving for $c$

Solving for $a$

Solving for $b^{2}$

$lim_{x \to 0} \frac{a x ^{2} e ^{x} - b l o g _{e} ( 1 + x ) + c x e ^{- x}}{x ^{2} s i n x} = 1$ then $16 (a^{2} + b^{2} + c^{2})$ is equal to _______.

If $lim_{x \to 0} \frac{a e ^{x} - b c o s x + c e ^{- x}}{x s i n x} = 2$ , then $a + b + c$ is equal to ______

If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then $2 α - β$ is equal to :

If $lim_{x \to 0} \frac{e ^{a x} - c o s ( b x ) - \frac{c x e ^{- c x}}{2}}{1 - c o s ( 2 x )} = 17$ , then $5 a^{2} + b^{2}$ is equal to

For $α, β, γ, \in R$ , if $lim_{x \to 0} \frac{x ^{2} s i n α x + ( γ - 1 ) e ^{x^{2}}}{s i n 2 x - β x} = 3$ , then $β + γ - α$ is equal to:

If $lim_{x \to 0} \frac{α x e ^{x} - β l o g _{e} ( 1 + x ) + γ x ^{2} e ^{- x}}{x s i n ^{2} x} = 10$ , $α, β, γ \in R$ , then the value of $α + β + γ$ is

If the value of $lim_{x \to 0} (2 - cos x cos 2 x)^{\frac{x + 2}{x ^{2}}}$ is equal to $e^{a}$ , then $a$ is equal to

If $lim_{x \to 0} \frac{α e ^{x} + β e ^{- x} + γ s i n x}{x s i n ^{2} x} = \frac{2}{3}$ , where $α, β, γ \in R$ , then which of the following is NOT correct ?

$lim_{x \to 0} (\frac{3 x ^{2} + 2}{7 x ^{2} + 2})^{\frac{1}{x ^{2}}}$ is equal to

If $lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}$ , then the values of $a$ and $b$ , are