JEE Main 2021 (18 March Shift 1)

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}$ is equal to $L$ , then the value of $(6 L + 1)$ is

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

Look at the denominator. The lowest power of $x$ in the denominator is $x^3$. You need to expand until you reach that power. If you stop too early, you get zero; if you go too far, you are just doing extra work.

Select Answer:

Visualized Solution

Identify the Limit Form

Given limit: $L = lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}$
Check the form by substituting $x = 0$ .
Numerator: $sin^{- 1} (0) - tan^{- 1} (0) = 0$
Denominator: $3 (0)^{3} = 0$
The limit is in the indeterminate form $\frac{0}{0}$ .

Series Expansion Method

Using L'Hopital's rule is too lengthy here.
We use Maclaurin series expansions:
$sin^{- 1} x = x + \frac{x ^{3}}{6} + \dots$
$tan^{- 1} x = x - \frac{x ^{3}}{3} + \dots$
We expand up to $x^{3}$ because the denominator is $3 x^{3}$ .

Substitute Expansions

Substitute the series into the limit expression:
$L = lim_{x \to 0} \frac{( x + \frac{x ^{3}}{6} + \dots ) - ( x - \frac{x ^{3}}{3} + \dots )}{3 x ^{3}}$

Simplify the Numerator

Open the brackets in the numerator:
$L = lim_{x \to 0} \frac{x - x + \frac{x ^{3}}{6} + \frac{x ^{3}}{3} + \dots}{3 x ^{3}}$
The linear terms cancel out: $x - x = 0$ .

Combine $x^{3}$ Terms

Combine the remaining $x^{3}$ terms:
$\frac{x ^{3}}{6} + \frac{x ^{3}}{3} = \frac{x ^{3} + 2 x ^{3}}{6}$
This simplifies to $\frac{3 x ^{3}}{6} = \frac{x ^{3}}{2}$ .

Evaluate the Limit $L$

The limit expression becomes:
$L = lim_{x \to 0} \frac{\frac{x ^{3}}{2}}{3 x ^{3}}$
Cancel $x^{3}$ from numerator and denominator:
$L = \frac{1}{2 \times 3} = \frac{1}{6}$

Final Calculation

We need to find the value of $(6 L + 1)$ .
Substitute $L = \frac{1}{6}$ :
$6 (\frac{1}{6}) + 1 = 1 + 1 = 2$
Final Answer: $2$

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

Analyzing the Setup

The problem asks us to evaluate the limit:

x \to 0 lim \frac{sin ^{- 1} x - tan ^{- 1} x}{3 x ^{3}}

While L'Hopital's rule is a valid approach, differentiating these inverse trigonometric functions three times leads to unnecessary algebraic complexity. The presence of

x^{3}

in the denominator is a strong indicator that Maclaurin series expansion is the most efficient strategy.

The Power of Series Expansion

We utilize the standard Maclaurin series expansions for

sin^{- 1} x

and

tan^{- 1} x

, truncating them at the

x^{3}

term since higher-order terms will approach zero as

x \to 0

sin^{- 1} x = x + \frac{x ^{3}}{6} + O (x^{5})

tan^{- 1} x = x - \frac{x ^{3}}{3} + O (x^{5})

The Algebraic Dance

Now, we substitute these expansions into the numerator of our limit expression:

Numerator = (x + \frac{x ^{3}}{6}) - (x - \frac{x ^{3}}{3})

Observe how the linear

x

terms cancel out perfectly:

Numerator = x - x + \frac{x ^{3}}{6} + \frac{x ^{3}}{3} = \frac{x ^{3}}{6} + \frac{2 x ^{3}}{6} = \frac{3 x ^{3}}{6} = \frac{x ^{3}}{2}

The entire numerator simplifies elegantly to

\frac{x ^{3}}{2}

The Final Victory

Substituting this back into the original limit, we have:

x \to 0 lim \frac{\frac{x ^{3}}{2}}{3 x ^{3}} = x \to 0 lim \frac{1}{2 \cdot 3} = \frac{1}{6}

Thus, we find

L = \frac{1}{6}

. The problem requires us to calculate

6 L + 1

6 (\frac{1}{6}) + 1 = 1 + 1 = 2

The final answer is 2.

Common Traps & FAQs

Why can't I just use L'Hopital's rule? It's a standard method, right?

It is a standard method, but it is a trap here. Differentiating

sin^{- 1} x

and

tan^{- 1} x

repeatedly leads to complex algebraic expressions that are prone to calculation errors. Series expansion is the elegant, 'JEE-smart' way to handle limits involving powers of

x

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

Look at the denominator. The lowest power of

x

in the denominator is

x^{3}

. You need to expand until you reach that power. If you stop too early, you get zero; if you go too far, you are just doing extra work.

Wait, why did the

x

terms cancel out? Did I make a mistake?

Not at all! That is exactly what should happen. Since the limit is

\frac{0}{0}

, the leading terms of the numerator must cancel out to reveal the higher-order behavior that determines the limit.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}$ is equal to $L$ , then the value of $(6 L + 1)$ is

Select Answer:

Visualized Solution

Identify the Limit Form

Series Expansion Method

Substitute Expansions

Simplify the Numerator

Combine $x^{3}$ Terms

Evaluate the Limit $L$

Final Calculation

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

Analyzing the Setup

The Power of Series Expansion

The Algebraic Dance

The Final Victory

Similar Questions

If $β = lim_{x \to 0} \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{1/3} + ( ( 1 - x ^{2} ) ^{1/2} - 1 ) s i n x}{x s i n ^{2} x}$ , then the value of $6 β$ is ______.

The value of $lim_{h \to 0} {\frac{3 s i n ( \frac{π}{6} + h ) - c o s ( \frac{π}{6} + h )}{3 h ( 3 c o s h - s i n h )}}$ is:

$lim_{x \to 0} \frac{x + 2 s i n x}{x ^{2} + 2 s i n x + 1 - s i n ^{2} x - x + 1}$ is :

$lim_{x \to \frac{1}{2}} \frac{s i n ( c o s ^{- 1} x ) - x}{1 - t a n ( c o s ^{- 1} x )}$ is equal to :

Let $f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t, x \in R$ . Then, $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ is equal to

The value of $lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{x ^{4} - 2 x ^{3} + 2 x - 1}$ is equal to:

$lim_{x \to \infty} \frac{( 3 x + 1 + 3 x - 1 ) ^{6} + ( 3 x + 1 - 3 x - 1 ) ^{6}}{( x + x ^{2} - 1 ) ^{6} + ( x - x ^{2} - 1 ) ^{6}} x^{3}$

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is

The value of $lim_{x \to 0} (\frac{x}{8 1 - s i n x - 8 1 + s i n x})$ is equal to:

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :

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; } If limx→0​3x3sin−1x−tan−1x​ is equal to L, then the value of (6L+1) is

Select Answer:

Visualized Solution

Identify the Limit Form

Series Expansion Method

Substitute Expansions

Simplify the Numerator

Combine .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x3 Terms

Evaluate the Limit .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Final Calculation

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

Analyzing the Setup

The Power of Series Expansion

The Algebraic Dance

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If β=limx→0​xsin2xex3−(1−x3)1/3+((1−x2)1/2−1)sinx​, then the value of 6β is ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limh→0​{3​h(3​cosh−sinh)3​sin(6π​+h)−cos(6π​+h)​} is:

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}$ is equal to $L$ , then the value of $(6 L + 1)$ is

Combine $x^{3}$ Terms

Evaluate the Limit $L$

If $β = lim_{x \to 0} \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{1/3} + ( ( 1 - x ^{2} ) ^{1/2} - 1 ) s i n x}{x s i n ^{2} x}$ , then the value of $6 β$ is ______.

The value of $lim_{h \to 0} {\frac{3 s i n ( \frac{π}{6} + h ) - c o s ( \frac{π}{6} + h )}{3 h ( 3 c o s h - s i n h )}}$ is:

$lim_{x \to 0} \frac{x + 2 s i n x}{x ^{2} + 2 s i n x + 1 - s i n ^{2} x - x + 1}$ is :

$lim_{x \to \frac{1}{2}} \frac{s i n ( c o s ^{- 1} x ) - x}{1 - t a n ( c o s ^{- 1} x )}$ is equal to :

Let $f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t, x \in R$ . Then, $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ is equal to

The value of $lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{x ^{4} - 2 x ^{3} + 2 x - 1}$ is equal to:

$lim_{x \to \infty} \frac{( 3 x + 1 + 3 x - 1 ) ^{6} + ( 3 x + 1 - 3 x - 1 ) ^{6}}{( x + x ^{2} - 1 ) ^{6} + ( x - x ^{2} - 1 ) ^{6}} x^{3}$

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is

The value of $lim_{x \to 0} (\frac{x}{8 1 - s i n x - 8 1 + s i n x})$ is equal to:

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :