JEE Main 2021 (20 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Q: Can I use Taylor Series expansion instead of L'Hospital's Rule?

Yes, and it is often faster! If you expand $\cos x \approx 1 - \frac{x^2}{2}$ and $\cos 2x \approx 1 - \frac{(2x)^2}{2}$, you can approximate the expression without differentiation. However, L'Hospital's Rule is a robust tool that works even when expansions become tedious. Both methods are valid paths to the same truth.

Q: Why does the $\sqrt{\cos 2x}$ in the denominator not cause a division by zero error?

Great observation! As $x \to 0$, $\cos 2x \to 1$, so $\sqrt{\cos 2x} \to 1$. Since the limit of the denominator is $1$ (which is non-zero), it does not affect the indeterminate nature of the limit. It is perfectly safe to evaluate that part of the limit separately.

Q: I got $a = 3/2$ instead of $3$. Where did I lose the factor of 2?

Don't forget the $(x+2)$ term in the exponent! When you split the limit, you have the core limit multiplied by $\lim_{x \to 0} (x+2)$. That limit is $2$. If you calculated the core limit as $3/2$, multiplying by $2$ gives you the correct answer of $3$. Always keep track of all factors!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Before diving into complex calculus, always test the limit. As

x \to 0

, observe the base and the exponent. Does it look like the classic

1^{\infty}

indeterminate form? If so, recall the identity

lim_{x \to a} [f (x)]^{g (x)} = e^{l i m_{x \to a} (f (x) - 1) g (x)}

Hint 2

Hint 2: Once you transform the limit into the exponent, you are left with

lim_{x \to 0} (1 - cos x cos 2 x) \frac{x + 2}{x ^{2}}

. Notice that

lim_{x \to 0} (x + 2)

is just a constant. Focus your energy on the core limit

lim_{x \to 0} \frac{1 - c o s x c o s 2 x}{x ^{2}}

Hint 3

Hint 3: To solve the core limit, you can use L'Hospital's Rule. When differentiating the numerator, remember the product rule and the chain rule carefully. After differentiation, look for the trigonometric identity

sin A cos B + cos A sin B = sin (A + B)

to simplify the expression into something manageable.

Visualized Solution

Identifying the Indeterminate Form

The given limit is $lim_{x \to 0} (2 - cos x cos 2 x)^{\frac{x + 2}{x ^{2}}}$ .
Check the form as $x \to 0$ :
Base: $2 - cos (0) cos (0) = 2 - 1 (1) = 1$ .
Exponent: $\frac{0 + 2}{0 ^{2}} \to \infty$ .
This is a $1^{\infty}$ indeterminate form.

Applying the $1^{\infty}$ Standard Formula

Using the property: $lim_{x \to a} [f (x)]^{g (x)} = e^{l i m_{x \to a} (f (x) - 1) g (x)}$ .
Here, $f (x) = 2 - cos x cos 2 x$ and $g (x) = \frac{x + 2}{x ^{2}}$ .

Setting up the Exponent Limit

So, the limit $L = e^{l i m_{x \to 0} (2 - c o s x c o s 2 x - 1) \frac{x + 2}{x ^{2}}}$ .
Simplifying the base term: $L = e^{l i m_{x \to 0} (1 - c o s x c o s 2 x) \frac{x + 2}{x ^{2}}}$ .

Separating the Limit

Separate the limit into two parts for easier evaluation:
$L = e^{l i m_{x \to 0} \frac{1 - c o s x c o s 2 x}{x ^{2}} \cdot l i m_{x \to 0} (x + 2)}$ .

Evaluating the Simple Limit

Evaluate the second part:
$lim_{x \to 0} (x + 2) = 0 + 2 = 2$ .

Analyzing the Core Limit

Consider the core limit: $lim_{x \to 0} \frac{1 - c o s x c o s 2 x}{x ^{2}}$ .
Substitute $x = 0$ : Numerator is $1 - 1 (1) = 0$ , Denominator is $0^{2} = 0$ .
This is a $\frac{0}{0}$ indeterminate form.

Applying L'Hospital's Rule

Apply L'Hospital's Rule: differentiate numerator and denominator.
$lim_{x \to 0} \frac{\frac{d}{d x} ( 1 - c o s x c o s 2 x )}{\frac{d}{d x} ( x ^{2} )}$ .
Denominator derivative: $\frac{d}{d x} (x^{2}) = 2 x$ .

Differentiating the Numerator

Numerator derivative: $\frac{d}{d x} (1 - cos x cos 2 x)$ .
Derivative of $1$ is $0$ .
Use product rule on $cos x \cdot (cos 2 x)^{\frac{1}{2}}$ :
$- [- sin x cos 2 x + cos x \cdot \frac{1}{2 c o s 2 x} (- sin 2 x \cdot 2)]$ .

Simplifying the Derivative

Simplify the expression:
$= sin x cos 2 x + \frac{c o s x s i n 2 x}{c o s 2 x}$ .
Take the Least Common Multiple (LCM):
$= \frac{s i n x ( c o s 2 x ) ^{2} + c o s x s i n 2 x}{c o s 2 x}$ .
$= \frac{s i n x c o s 2 x + c o s x s i n 2 x}{c o s 2 x}$ .

Using the Sine Addition Identity

Notice the numerator: $sin x cos 2 x + cos x sin 2 x$ .
This matches the identity: $sin (A + B) = sin A cos B + cos A sin B$ .
Here $A = x$ and $B = 2 x$ .
Numerator becomes: $sin (x + 2 x) = sin 3 x$ .
So, the derivative is $\frac{s i n 3 x}{c o s 2 x}$ .

Reassembling the Core Limit

Substitute the simplified derivative back into the limit:
$lim_{x \to 0} \frac{\frac{s i n 3 x}{c o s 2 x}}{2 x}$ .
Rearrange the terms:
$= lim_{x \to 0} \frac{s i n 3 x}{2 x c o s 2 x}$ .

Evaluating the Final Limit

Rewrite to use the standard limit $lim_{θ \to 0} \frac{s i n θ}{θ} = 1$ :
$= \frac{1}{2} \cdot lim_{x \to 0} \frac{s i n 3 x}{x} \cdot lim_{x \to 0} \frac{1}{c o s 2 x}$ .
Multiply and divide by $3$ :
$= \frac{3}{2} \cdot lim_{x \to 0} \frac{s i n 3 x}{3 x} \cdot lim_{x \to 0} \frac{1}{c o s 2 x}$ .
$= \frac{3}{2} \cdot 1 \cdot \frac{1}{1} = \frac{3}{2}$ .

Combining and Concluding

Recall the original limit structure: $L = e^{L_{1} \cdot L_{2}}$ where $L_{1}$ is the core limit and $L_{2}$ is the simple limit.
Core Limit $L_{1} = \frac{3}{2}$ .
Simple Limit $L_{2} = 2$ .
$L = e^{\frac{3}{2} \cdot 2} = e^{3}$ .
Given $L = e^{a}$ , comparing gives $a = 3$ .

00:00 / 00:00

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

Analyzing the Setup

We are evaluating the limit:

x \to 0 lim (2 - cos x cos 2 x)^{\frac{x + 2}{x ^{2}}}

When

x \to 0

, the base approaches

2 - cos (0) cos (0) = 2 - 1 (1) = 1

. Simultaneously, the exponent

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

approaches

\infty

We have identified the classic $1^{\infty}$ indeterminate form. This indicates that we must use the standard exponential transformation to resolve the limit.

The Elegant Transformation

To resolve this, we utilize the identity:

x \to a lim [f (x)]^{g (x)} = e^{l i m_{x \to a} (f (x) - 1) g (x)}

Applying this to our expression, the limit

L

becomes:

L = e^{l i m_{x \to 0} (2 - c o s x c o s 2 x - 1) \frac{x + 2}{x ^{2}}}

Simplifying the base term, we obtain:

L = e^{l i m_{x \to 0} (1 - c o s x c o s 2 x) \frac{x + 2}{x ^{2}}}

We can split this into two parts: a core limit and a simple limit. Since

lim_{x \to 0} (x + 2) = 2

, the expression simplifies to:

L = e^{2 \cdot l i m_{x \to 0} \frac{1 - c o s x c o s 2 x}{x ^{2}}}

The Calculus of Simplification

We now focus on the core limit:

x \to 0 lim \frac{1 - cos x cos 2 x}{x ^{2}}

Substituting

x = 0

yields the

\frac{0}{0}

form, allowing us to apply L'Hospital's Rule. Differentiating the denominator gives

2 x

Differentiating the numerator

1 - cos x cos 2 x

using the product and chain rules yields:

- [- sin x cos 2 x + cos x \cdot \frac{1}{2 cos 2 x} \cdot (- sin 2 x \cdot 2)]

Simplifying the expression inside the bracket, we get:

\frac{sin x cos 2 x + cos x sin 2 x}{cos 2 x}

The Trigonometric Collapse

The numerator

sin x cos 2 x + cos x sin 2 x

is the expansion of the sine addition identity

sin (A + B)

, where

A = x

and

B = 2 x

. Thus, the numerator collapses into

sin 3 x

The core limit now becomes:

x \to 0 lim \frac{sin 3 x}{2 x cos 2 x}

We can rewrite this as:

\frac{3}{2} \cdot x \to 0 lim (\frac{sin 3 x}{3 x}) \cdot x \to 0 lim (\frac{1}{cos 2 x})

Since

lim_{θ \to 0} \frac{s i n θ}{θ} = 1

and

\frac{1}{c o s ( 0 )} = 1

, the core limit evaluates to

\frac{3}{2}

Final Calculation

Returning to our original structure for

L

L = e^{2 \cdot \frac{3}{2}} = e^{3}

The final value of the limit is $e^{3}$ .

Common Traps & FAQs

Wait, why do we use the

e^{l i m (f - 1) g}

formula? Can't we just take the natural log of both sides?

You absolutely can! Taking the natural log is the fundamental derivation of that formula. By setting

y = f (x)^{g (x)}

, then

ln y = g (x) ln (f (x))

. As

x \to 0

ln y \to \infty \cdot ln (1) = \infty \cdot 0

. Using the expansion

ln (1 + u) \approx u

, you arrive at the same result. The formula is just a shortcut for this exact process!

Why did the derivative of the numerator look so messy? Did I do something wrong?

Calculus often gets messy before it gets beautiful! When you differentiate

cos x cos 2 x

, the product rule creates two terms. Once you find a common denominator, those terms often combine into a standard identity. Trust the process—the complexity usually collapses once you apply the sine addition formula.

Can I use Taylor Series expansion instead of L'Hospital's Rule?

Yes, and it is often faster! If you expand

cos x \approx 1 - \frac{x ^{2}}{2}

and

cos 2 x \approx 1 - \frac{( 2 x ) ^{2}}{2}

, you can approximate the expression without differentiation. However, L'Hospital's Rule is a robust tool that works even when expansions become tedious. Both methods are valid paths to the same truth.

Why does the

cos 2 x

in the denominator not cause a division by zero error?

Great observation! As

x \to 0

cos 2 x \to 1

, so

cos 2 x \to 1

. Since the limit of the denominator is

1

(which is non-zero), it does not affect the indeterminate nature of the limit. It is perfectly safe to evaluate that part of the limit separately.

I got

a = 3/2

instead of

3

. Where did I lose the factor of 2?

Don't forget the

(x + 2)

term in the exponent! When you split the limit, you have the core limit multiplied by

lim_{x \to 0} (x + 2)

. That limit is

2

. If you calculated the core limit as

3/2

, multiplying by

2

gives you the correct answer of

3

. Always keep track of all factors!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Enter Numerical Value:

Visualized Solution

Identifying the Indeterminate Form

Applying the $1^{\infty}$ Standard Formula

Setting up the Exponent Limit

Separating the Limit

Evaluating the Simple Limit

Analyzing the Core Limit

Applying L'Hospital's Rule

Differentiating the Numerator

Simplifying the Derivative

Using the Sine Addition Identity

Reassembling the Core Limit

Evaluating the Final Limit

Combining and Concluding

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

Analyzing the Setup

The Elegant Transformation

The Calculus of Simplification

The Trigonometric Collapse

Final Calculation

Similar Questions

Evaluate $lim_{x \to a} \frac{a + 2 x - 3 x}{3 a + x - 2 x}, (a \neq = 0) .$

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

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

$lim_{x \to a} \frac{( a + 2 x ) ^{1/3} - ( 3 x ) ^{1/3}}{( 3 a + x ) ^{1/3} - ( 4 x ) ^{1/3}}$ $(a \neq = 0)$ is equal to

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

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 :

The value of $lim_{x \to 0} 2 (\frac{1 - c o s x c o s 2 x 3 c o s 3 x \dots 10 c o s 10 x}{x ^{2}})$ is

If $lim_{x \to 0} \frac{a x - ( e ^{4 x} - 1 )}{a x ( e ^{4 x} - 1 )}$ exists and is equal to $b$ , then the value of $a - 2 b$ is

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

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 the value of limx→0​(2−cosxcos2x​)x2x+2​ is equal to ea, then a is equal to

Enter Numerical Value:

Visualized Solution

Identifying the Indeterminate Form

Applying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1∞ Standard Formula

Setting up the Exponent Limit

Separating the Limit

Evaluating the Simple Limit

Analyzing the Core Limit

Applying L'Hospital's Rule

Differentiating the Numerator

Simplifying the Derivative

Using the Sine Addition Identity

Reassembling the Core Limit

Evaluating the Final Limit

Combining and Concluding

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

Analyzing the Setup

The Elegant Transformation

The Calculus of Simplification

The Trigonometric Collapse

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Applying the $1^{\infty}$ Standard Formula

Evaluate $lim_{x \to a} \frac{a + 2 x - 3 x}{3 a + x - 2 x}, (a \neq = 0) .$

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

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

$lim_{x \to a} \frac{( a + 2 x ) ^{1/3} - ( 3 x ) ^{1/3}}{( 3 a + x ) ^{1/3} - ( 4 x ) ^{1/3}}$ $(a \neq = 0)$ is equal to

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

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 :

The value of $lim_{x \to 0} 2 (\frac{1 - c o s x c o s 2 x 3 c o s 3 x \dots 10 c o s 10 x}{x ^{2}})$ is

If $lim_{x \to 0} \frac{a x - ( e ^{4 x} - 1 )}{a x ( e ^{4 x} - 1 )}$ exists and is equal to $b$ , then the value of $a - 2 b$ is

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