Let $L = \lim_{x \to 0} \frac{a - \sqrt{a^2 - x^2} - \frac{x^2}{4}}{x^4}, a > 0$. If $L$ is finite, then Options: (A) $a = 2$ (B) $a = 1$ (C) $L = 1/64$ (D) $L = 1/32$

### Analyzing the Setup We are tasked with evaluating the limit: $$L = \lim_{x \to 0} \frac{a - \sqrt{a^2 - x^2} - \frac{x^2}{4}}{x^4}$$ As $x \to 0$, the denominator $x^4$ approaches zero. For the limit $L$ to be finite, the numerator must also approach zero, creating an indeterminate form of $\frac{0}{0}$. This requirement forces the higher-order terms of the numerator to cancel out perfectly. ### The Binomial Weapon To analyze the behavior of the square root near $x=0$, we utilize the **Binomial Theorem**. We rewrite the expression as follows: $$\sqrt{a^2 - x^2} = a \left( 1 - \frac{x^2}{a^2} \right)^{1/2}$$ Using the expansion $(1-z)^n = 1 - nz + \frac{n(n-1)}{2!}z^2 - \dots$, we expand the term up to the $x^4$ power. This precision is necessary because the denominator is $x^4$. The expansion yields: $$\sqrt{a^2 - x^2} = a \left[ 1 - \frac{1}{2}\left(\frac{x^2}{a^2}\right) - \frac{1}{8}\left(\frac{x^2}{a^2}\right)^2 - \dots \right]$$ Distributing the $a$, we obtain: $$\sqrt{a^2 - x^2} = a - \frac{x^2}{2a} - \frac{x^4}{8a^3}$$ ### The Balancing Act Substituting this expansion back into the original numerator, we get: $$Numerator = a - \left( a - \frac{x^2}{2a} - \frac{x^4}{8a^3} \right) - \frac{x^2}{4}$$ Simplifying the expression by grouping terms of $x^2$ and $x^4$: $$Numerator = x^2 \left( \frac{1}{2a} - \frac{1}{4} \right) + \frac{x^4}{8a^3}$$ For the limit to be finite, the coefficient of the $x^2$ term must be zero. If it were non-zero, the limit would involve a term of $\frac{1}{x^2}$, which diverges as $x \to 0$. Setting the coefficient to zero: $$\frac{1}{2a} - \frac{1}{4} = 0 \Rightarrow a = 2$$ ### Final Calculation With $a=2$, the $x^2$ term vanishes, leaving us with the $x^4$ terms: $$L = \lim_{x \to 0} \frac{\frac{x^4}{8a^3}}{x^4} = \frac{1}{8a^3}$$ Substituting $a=2$ into this result: $$L = \frac{1}{8(2)^3} = \frac{1}{8 \times 8} = \frac{1}{64}$$ The final value of the limit is **$\frac{1}{64}$**.

JEE Advanced 2009

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $L = lim_{x \to 0} \frac{a - a ^{2} - x ^{2} - \frac{x ^{2}}{4}}{x ^{4}}, a > 0$ . If $L$ is finite, then

Q: Why do we expand the square root only up to the $x^4$ term? Why not further?

Great question! We expand up to $x^4$ because the denominator is $x^4$. If we expanded further (to $x^6$, etc.), those higher-order terms would contain $x^2, x^4, \dots$ in the numerator after division, which would all approach zero as $x \to 0$. Expanding up to the degree of the denominator is sufficient to capture the non-zero limit.

Q: Wait, why must the coefficient of the $x^2$ term be zero? Can't it just be a constant?

If the coefficient of $x^2$ were a non-zero constant, say $C$, then the term in the limit would look like $\frac{Cx^2}{x^4} = \frac{C}{x^2}$. As $x \to 0$, $\frac{C}{x^2}$ approaches infinity. Since the problem explicitly states that the limit $L$ is finite, we are forced to make that $x^2$ term vanish by setting its coefficient to zero.

Select Answer:

* Multiple Correct

Visualized Solution

Understanding the Limit Problem

Given limit: $L = lim_{x \to 0} \frac{a - a ^{2} - x ^{2} - \frac{x ^{2}}{4}}{x ^{4}}$
Constraint: $a > 0$ and $L$ is finite.
Objective: Find the value of $a$ and the resulting limit $L$ .

Analyzing the Indeterminate Form

As $x \to 0$ , the denominator $x^{4} \to 0$ .
For the limit to be finite, the numerator must also approach $0$ .
The presence of $a^{2} - x^{2}$ suggests using the Binomial Expansion for fractional powers.

Preparing the Square Root Term

Rewrite the square root: $a^{2} - x^{2} = (a^{2} - x^{2})^{\frac{1}{2}}$
Factor out $a^{2}$ : $(a^{2} (1 - \frac{x ^{2}}{a ^{2}}))^{\frac{1}{2}}$
Simplify: $a (1 - \frac{x ^{2}}{a ^{2}})^{\frac{1}{2}}$ (since $a > 0$ )

Binomial Expansion of $(1 - z)^{n}$

Standard formula: $(1 - z)^{n} = 1 - n z + \frac{n ( n - 1 )}{2 !} z^{2} - \dots$
Here, $z = \frac{x ^{2}}{a ^{2}}$ and $n = \frac{1}{2}$ .
Expand up to $z^{2}$ (which corresponds to $x^{4}$ ):
$(1 - \frac{x ^{2}}{a ^{2}})^{\frac{1}{2}} \approx 1 - \frac{1}{2} (\frac{x ^{2}}{a ^{2}}) + \frac{\frac{1}{2} ( \frac{1}{2} - 1 )}{2} (\frac{x ^{2}}{a ^{2}})^{2}$

Simplifying the Expanded Term

Simplify the coefficients:
$\frac{\frac{1}{2} ( - \frac{1}{2} )}{2} = - \frac{1}{8}$
The expansion becomes: $1 - \frac{x ^{2}}{2 a ^{2}} - \frac{x ^{4}}{8 a ^{4}}$
Multiply by $a$ : $a^{2} - x^{2} \approx a - \frac{x ^{2}}{2 a} - \frac{x ^{4}}{8 a ^{3}}$

Reconstructing the Numerator

Original Numerator: $a - a^{2} - x^{2} - \frac{x ^{2}}{4}$
Substitute the expansion:
$a - (a - \frac{x ^{2}}{2 a} - \frac{x ^{4}}{8 a ^{3}}) - \frac{x ^{2}}{4}$
Distribute the negative sign:
$a - a + \frac{x ^{2}}{2 a} + \frac{x ^{4}}{8 a ^{3}} - \frac{x ^{2}}{4}$

Grouping Powers of $x$

Cancel $a - a = 0$ .
Group $x^{2}$ terms: $x^{2} (\frac{1}{2 a} - \frac{1}{4})$
Remaining $x^{4}$ term: $\frac{x ^{4}}{8 a ^{3}}$
Simplified Numerator: $x^{2} (\frac{1}{2 a} - \frac{1}{4}) + \frac{x ^{4}}{8 a ^{3}}$

Condition for a Finite Limit

The limit is $L = lim_{x \to 0} \frac{x ^{2} ( \frac{1}{2 a} - \frac{1}{4} ) + \frac{x ^{4}}{8 a ^{3}}}{x ^{4}}$
Separate the fraction: $lim_{x \to 0} [\frac{1}{x ^{2}} (\frac{1}{2 a} - \frac{1}{4}) + \frac{1}{8 a ^{3}}]$
For $L$ to be finite, the term with $\frac{1}{x ^{2}}$ must vanish.
Therefore, the coefficient of $x^{2}$ must be zero: $\frac{1}{2 a} - \frac{1}{4} = 0$

Finding the Value of $a$

Set the coefficient to zero: $\frac{1}{2 a} - \frac{1}{4} = 0$
Rearrange: $\frac{1}{2 a} = \frac{1}{4}$
Cross-multiply: $2 a = 4$
Solve: $a = 2$

Calculating the Limit $L$

With the $x^{2}$ term gone, the limit simplifies to:
$L = lim_{x \to 0} \frac{\frac{x ^{4}}{8 a ^{3}}}{x ^{4}}$
Cancel $x^{4}$ : $L = \frac{1}{8 a ^{3}}$
Substitute $a = 2$ : $L = \frac{1}{8 ( 2 ) ^{3}}$
Calculate: $L = \frac{1}{8 \cdot 8} = \frac{1}{64}$

Final Results

The value of the constant is $a = 2$ .
The finite limit evaluates to $L = \frac{1}{64}$ .
Both conditions are satisfied, confirming our solution.

00:00 / 00:00

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

Analyzing the Setup

We are tasked with evaluating the limit:

L = x \to 0 lim \frac{a - a ^{2} - x ^{2} - \frac{x ^{2}}{4}}{x ^{4}}

x \to 0

, the denominator

x^{4}

approaches zero. For the limit

L

to be finite, the numerator must also approach zero, creating an indeterminate form of

\frac{0}{0}

. This requirement forces the higher-order terms of the numerator to cancel out perfectly.

The Binomial Weapon

To analyze the behavior of the square root near

x = 0

, we utilize the Binomial Theorem. We rewrite the expression as follows:

a^{2} - x^{2} = a (1 - \frac{x ^{2}}{a ^{2}})^{1/2}

Using the expansion

(1 - z)^{n} = 1 - n z + \frac{n ( n - 1 )}{2 !} z^{2} - \dots

, we expand the term up to the

x^{4}

power. This precision is necessary because the denominator is

x^{4}

The expansion yields:

a^{2} - x^{2} = a [1 - \frac{1}{2} (\frac{x ^{2}}{a ^{2}}) - \frac{1}{8} (\frac{x ^{2}}{a ^{2}})^{2} - \dots]

Distributing the

a

, we obtain:

a^{2} - x^{2} = a - \frac{x ^{2}}{2 a} - \frac{x ^{4}}{8 a ^{3}}

The Balancing Act

Substituting this expansion back into the original numerator, we get:

N u m er a t or = a - (a - \frac{x ^{2}}{2 a} - \frac{x ^{4}}{8 a ^{3}}) - \frac{x ^{2}}{4}

Simplifying the expression by grouping terms of

x^{2}

and

x^{4}

N u m er a t or = x^{2} (\frac{1}{2 a} - \frac{1}{4}) + \frac{x ^{4}}{8 a ^{3}}

For the limit to be finite, the coefficient of the

x^{2}

term must be zero. If it were non-zero, the limit would involve a term of

\frac{1}{x ^{2}}

, which diverges as

x \to 0

. Setting the coefficient to zero:

\frac{1}{2 a} - \frac{1}{4} = 0 \Rightarrow a = 2

Final Calculation

With

a = 2

, the

x^{2}

term vanishes, leaving us with the

x^{4}

terms:

L = x \to 0 lim \frac{\frac{x ^{4}}{8 a ^{3}}}{x ^{4}} = \frac{1}{8 a ^{3}}

Substituting

a = 2

into this result:

L = \frac{1}{8 ( 2 ) ^{3}} = \frac{1}{8 \times 8} = \frac{1}{64}

The final value of the limit is $\frac{1}{64}$ .

Common Traps & FAQs

Why do we expand the square root only up to the

x^{4}

term? Why not further?

Great question! We expand up to

x^{4}

because the denominator is

x^{4}

. If we expanded further (to

x^{6}

, etc.), those higher-order terms would contain

x^{2}, x^{4}, \dots

in the numerator after division, which would all approach zero as

x \to 0

. Expanding up to the degree of the denominator is sufficient to capture the non-zero limit.

Wait, why must the coefficient of the

x^{2}

term be zero? Can't it just be a constant?

If the coefficient of

x^{2}

were a non-zero constant, say

C

, then the term in the limit would look like

\frac{C x ^{2}}{x ^{4}} = \frac{C}{x ^{2}}

. As

x \to 0

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

approaches infinity. Since the problem explicitly states that the limit

L

is finite, we are forced to make that

x^{2}

term vanish by setting its coefficient to zero.

Can we use L'Hopital's Rule instead of binomial expansion?

Yes, you absolutely can! However, applying L'Hopital's Rule four times on a square root function can become algebraically messy and prone to calculation errors. Binomial expansion is generally more elegant and faster for limits involving roots or powers as

x \to 0

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $L = lim_{x \to 0} \frac{a - a ^{2} - x ^{2} - \frac{x ^{2}}{4}}{x ^{4}}, a > 0$ . If $L$ is finite, then

Select Answer:

Visualized Solution

Understanding the Limit Problem

Analyzing the Indeterminate Form

Preparing the Square Root Term

Binomial Expansion of $(1 - z)^{n}$

Simplifying the Expanded Term

Reconstructing the Numerator

Grouping Powers of $x$

Condition for a Finite Limit

Finding the Value of $a$

Calculating the Limit $L$

Final Results

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

Analyzing the Setup

The Binomial Weapon

The Balancing Act

Final Calculation

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let L=limx→0​x4a−a2−x2​−4x2​​,a>0. If L is finite, then

Select Answer:

Visualized Solution

Understanding the Limit Problem

Analyzing the Indeterminate Form

Preparing the Square Root Term

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