$\lim_{x \to 0} \left( \frac{(x + 2\cos x)^3 + 2(x + 2\cos x)^2 + 3\sin(x + 2\cos x)}{(x+2)^3 + 2(x+2)^2 + 3\sin(x+2)} \right)^{\frac{100}{x}}$ is equal to ______.

### Analyzing the Setup The expression we are evaluating is: $$\lim_{x \to 0} \left( \frac{(x + 2\cos x)^3 + 2(x + 2\cos x)^2 + 3\sin(x + 2\cos x)}{(x+2)^3 + 2(x+2)^2 + 3\sin(x+2)} \right)^{\frac{100}{x}}$$ To begin, we test the limit as $x \to 0$. In the numerator, the term $(x + 2\cos x)$ approaches $0 + 2(1) = 2$. In the denominator, $(x + 2)$ also approaches $0 + 2 = 2$. Both the numerator and denominator evaluate to $2^3 + 2(2)^2 + 3\sin(2) = 16 + 3\sin 2$. Since the base approaches 1 and the exponent $\frac{100}{x}$ approaches $\infty$, we have identified the classic **$1^\infty$ indeterminate form**. ### The Master Equation To resolve this, we utilize the exponential limit identity: $$\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)(f(x) - 1)}$$ We define our limit as $e^L$, where $L = \lim_{x \to 0} \frac{100}{x} \left( \frac{f(x)}{g(x)} - 1 \right)$. By finding a common denominator, this simplifies to: $$L = \lim_{x \to 0} \frac{100}{x} \cdot \frac{f(x) - g(x)}{g(x)}$$ Since $g(0) = 16 + 3\sin 2$ is a non-zero constant, we can extract it from the limit: $$L = \frac{100}{g(0)} \lim_{x \to 0} \frac{f(x) - g(x)}{x}$$ ### The Beauty of the Derivative Definition Because $f(0) = g(0)$, we can rewrite the numerator as $(f(x) - f(0)) - (g(x) - g(0))$. The limit then transforms into the definition of the derivative: $$\lim_{x \to 0} \frac{f(x) - f(0)}{x} - \lim_{x \to 0} \frac{g(x) - g(0)}{x} = f'(0) - g'(0)$$ This reduces the complex expression to a simple subtraction of two derivatives. ### Final Calculation Let $f(x) = (x + 2\cos x)^3 + 2(x + 2\cos x)^2 + 3\sin(x + 2\cos x)$. Using the chain rule with $u = x + 2\cos x$, we find: $$f'(x) = [3u^2 + 4u + 3\cos(u)] \cdot (1 - 2\sin x)$$ At $x=0$, $u=2$, so $f'(0) = [3(2)^2 + 4(2) + 3\cos(2)] \cdot (1) = 20 + 3\cos 2$. Now, for $g(x) = (x + 2)^3 + 2(x + 2)^2 + 3\sin(x + 2)$, the derivative is: $$g'(x) = 3(x + 2)^2 + 4(x + 2) + 3\cos(x + 2)$$ At $x=0$, $g'(0) = 3(2)^2 + 4(2) + 3\cos(2) = 20 + 3\cos 2$. The difference $f'(0) - g'(0)$ is exactly $0$. Thus, $L = 0$, and our final answer is: **$e^0 = 1$**

JEE Main 2022 (28 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to 0} (\frac{( x + 2 c o s x ) ^{3} + 2 ( x + 2 c o s x ) ^{2} + 3 s i n ( x + 2 c o s x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 s i n ( x + 2 )})^{\frac{100}{x}}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyze the Limit Structure

Given limit: $lim_{x \to 0} (\frac{( x + 2 c o s x ) ^{3} + 2 ( x + 2 c o s x ) ^{2} + 3 s i n ( x + 2 c o s x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 s i n ( x + 2 )})^{\frac{100}{x}}$
Let the base be $H (x) = \frac{f ( x )}{g ( x )}$ and the exponent be $h (x) = \frac{100}{x}$ .

Evaluate the Base at $x = 0$

As $x \to 0$ , $(x + 2 cos x) \to (0 + 2 (1)) = 2$ .
As $x \to 0$ , $(x + 2) \to (0 + 2) = 2$ .
Numerator at $x = 0$ : $2^{3} + 2 (2)^{2} + 3 sin (2) = 16 + 3 sin 2$ .
Denominator at $x = 0$ : $2^{3} + 2 (2)^{2} + 3 sin (2) = 16 + 3 sin 2$ .
Thus, the base $\to 1$ as $x \to 0$ .

Identify the $1^{\infty}$ Form

Base $\to 1$ and Exponent $\frac{100}{x} \to \infty$ as $x \to 0$ .
This is the indeterminate form $1^{\infty}$ .
Standard Formula: $lim_{x \to a} [f (x)]^{g (x)} = e^{l i m_{x \to a} g (x) (f (x) - 1)}$ if $lim f (x) = 1$ and $lim g (x) = \infty$ .

Apply the Exponential Transformation

Let the limit be $e^{L}$ .
$L = lim_{x \to 0} \frac{100}{x} (\frac{f ( x )}{g ( x )} - 1)$
Where $f (x) = (x + 2 cos x)^{3} + 2 (x + 2 cos x)^{2} + 3 sin (x + 2 cos x)$
And $g (x) = (x + 2)^{3} + 2 (x + 2)^{2} + 3 sin (x + 2)$

Simplify the Exponent Expression

$L = lim_{x \to 0} \frac{100}{x} \cdot \frac{f ( x ) - g ( x )}{g ( x )}$
Since $g (0) = 16 + 3 sin 2 \neq = 0$ , we can write:
$L = \frac{100}{g ( 0 )} lim_{x \to 0} \frac{f ( x ) - g ( x )}{x}$

Relate to the Derivative Definition

Note that $f (0) = g (0)$ .
$lim_{x \to 0} \frac{f ( x ) - g ( x )}{x} = lim_{x \to 0} (\frac{f ( x ) - f ( 0 )}{x} - \frac{g ( x ) - g ( 0 )}{x})$
By definition of derivative: $L = \frac{100}{g ( 0 )} (f^{'} (0) - g^{'} (0))$

Differentiate $f (x)$ using Chain Rule

$f (x) = (x + 2 cos x)^{3} + 2 (x + 2 cos x)^{2} + 3 sin (x + 2 cos x)$
Let $u = x + 2 cos x$ , then $\frac{d u}{d x} = 1 - 2 sin x$ .
$f^{'} (x) = [3 u^{2} + 4 u + 3 cos (u)] \cdot \frac{d u}{d x}$
$f^{'} (x) = [3 (x + 2 cos x)^{2} + 4 (x + 2 cos x) + 3 cos (x + 2 cos x)] (1 - 2 sin x)$

Calculate $f^{'} (0)$

At $x = 0$ , $u = 0 + 2 (1) = 2$ and $\frac{d u}{d x} = 1 - 2 (0) = 1$ .
$f^{'} (0) = [3 (2)^{2} + 4 (2) + 3 cos (2)] \cdot (1)$
$f^{'} (0) = 12 + 8 + 3 cos 2 = 20 + 3 cos 2$

Differentiate $g (x)$ and Calculate $g^{'} (0)$

$g (x) = (x + 2)^{3} + 2 (x + 2)^{2} + 3 sin (x + 2)$
$g^{'} (x) = 3 (x + 2)^{2} + 4 (x + 2) + 3 cos (x + 2)$
$g^{'} (0) = 3 (2)^{2} + 4 (2) + 3 cos (2) = 20 + 3 cos 2$

Final Calculation and Conclusion

We found $f^{'} (0) = 20 + 3 cos 2$ and $g^{'} (0) = 20 + 3 cos 2$ .
Therefore, $f^{'} (0) - g^{'} (0) = 0$ .
Substitute back into $L$ : $L = \frac{100}{g ( 0 )} (0) = 0$ .
Final Limit $= e^{L} = e^{0} = 1$ .
Key Takeaway: For $1^{\infty}$ forms involving complex functions, look for derivative patterns to simplify the limit.

00:00 / 00:00

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

Analyzing the Setup

The expression we are evaluating is:

x \to 0 lim (\frac{( x + 2 cos x ) ^{3} + 2 ( x + 2 cos x ) ^{2} + 3 sin ( x + 2 cos x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 sin ( x + 2 )})^{\frac{100}{x}}

To begin, we test the limit as

x \to 0

. In the numerator, the term

(x + 2 cos x)

approaches

0 + 2 (1) = 2

In the denominator,

(x + 2)

also approaches

0 + 2 = 2

. Both the numerator and denominator evaluate to

2^{3} + 2 (2)^{2} + 3 sin (2) = 16 + 3 sin 2

Since the base approaches 1 and the exponent

\frac{100}{x}

approaches

\infty

, we have identified the classic $1^{\infty}$ indeterminate form.

The Master Equation

To resolve this, we utilize the exponential limit identity:

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

We define our limit as

e^{L}

, where

L = lim_{x \to 0} \frac{100}{x} (\frac{f ( x )}{g ( x )} - 1)

. By finding a common denominator, this simplifies to:

L = x \to 0 lim \frac{100}{x} \cdot \frac{f ( x ) - g ( x )}{g ( x )}

Since

g (0) = 16 + 3 sin 2

is a non-zero constant, we can extract it from the limit:

L = \frac{100}{g ( 0 )} x \to 0 lim \frac{f ( x ) - g ( x )}{x}

The Beauty of the Derivative Definition

Because

f (0) = g (0)

, we can rewrite the numerator as

(f (x) - f (0)) - (g (x) - g (0))

. The limit then transforms into the definition of the derivative:

x \to 0 lim \frac{f ( x ) - f ( 0 )}{x} - x \to 0 lim \frac{g ( x ) - g ( 0 )}{x} = f^{'} (0) - g^{'} (0)

This reduces the complex expression to a simple subtraction of two derivatives.

Final Calculation

Let

f (x) = (x + 2 cos x)^{3} + 2 (x + 2 cos x)^{2} + 3 sin (x + 2 cos x)

. Using the chain rule with

u = x + 2 cos x

, we find:

f^{'} (x) = [3 u^{2} + 4 u + 3 cos (u)] \cdot (1 - 2 sin x)

x = 0

u = 2

, so

f^{'} (0) = [3 (2)^{2} + 4 (2) + 3 cos (2)] \cdot (1) = 20 + 3 cos 2

Now, for

g (x) = (x + 2)^{3} + 2 (x + 2)^{2} + 3 sin (x + 2)

, the derivative is:

g^{'} (x) = 3 (x + 2)^{2} + 4 (x + 2) + 3 cos (x + 2)

x = 0

g^{'} (0) = 3 (2)^{2} + 4 (2) + 3 cos (2) = 20 + 3 cos 2

The difference

f^{'} (0) - g^{'} (0)

is exactly

0

. Thus,

L = 0

, and our final answer is:

$e^{0} = 1$

Common Traps & FAQs

Wait, why do we reject the negative root here? Or rather, why is the base approaching 1?

Great question! We aren't rejecting anything. We are simply evaluating the limit of the base as

x

approaches 0. By substituting

x = 0

into the expression

(x + 2 cos x)

, we get

0 + 2 (1) = 2

. Plugging

x = 0

into the denominator

(x + 2)

also gives 2. Since the numerator and denominator become identical at

x = 0

, the ratio is 1. It is not a root-finding problem; it is a direct substitution to identify the indeterminate form.

Why can't we just use L'Hopital's rule on the whole thing from the start?

You could, but it would be a nightmare! Differentiating that massive expression directly would lead to an algebraic disaster. The beauty of the

1^{\infty}

formula is that it transforms the problem into a derivative difference,

f^{'} (0) - g^{'} (0)

, which is much cleaner. Always look for the most elegant path before diving into brute-force differentiation.

How did you know that

f (0) = g (0)

That is the key observation! By calculating the value of the numerator and denominator at

x = 0

, we found they both equal

16 + 3 sin 2

. Recognizing this equality is exactly what allows us to use the definition of the derivative,

lim_{x \to 0} \frac{f ( x ) - f ( 0 )}{x} - \frac{g ( x ) - g ( 0 )}{x}

, to solve the problem efficiently.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to 0} (\frac{( x + 2 c o s x ) ^{3} + 2 ( x + 2 c o s x ) ^{2} + 3 s i n ( x + 2 c o s x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 s i n ( x + 2 )})^{\frac{100}{x}}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyze the Limit Structure

Evaluate the Base at $x = 0$

Identify the $1^{\infty}$ Form

Apply the Exponential Transformation

Simplify the Exponent Expression

Relate to the Derivative Definition

Differentiate $f (x)$ using Chain Rule

Calculate $f^{'} (0)$

Differentiate $g (x)$ and Calculate $g^{'} (0)$

Final Calculation and Conclusion

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

Analyzing the Setup

The Master Equation

The Beauty of the Derivative Definition

Final Calculation

Similar Questions

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

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

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

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

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ 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 ______

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ equals

$lim_{x \to 0} csc x (2 cos^{2} x + 3 cos x - cos^{2} x + sin x + 4)$ is:

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to π /4} \frac{c o t ^{3} x - t a n x}{c o s ( x + π /4 )}$ is :

Enter Numerical Value:

Visualized Solution

Analyze the Limit Structure

Evaluate the Base at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

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

Apply the Exponential Transformation

Simplify the Exponent Expression

Relate to the Derivative Definition

Differentiate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) using Chain Rule

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(0)

Differentiate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) and Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(0)

Final Calculation and Conclusion

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

Analyzing the Setup

The Master Equation

The Beauty of the Derivative Definition

Final Calculation

Similar Questions

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Evaluate the Base at $x = 0$

Identify the $1^{\infty}$ Form

Differentiate $f (x)$ using Chain Rule

Calculate $f^{'} (0)$

Differentiate $g (x)$ and Calculate $g^{'} (0)$

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

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

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

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

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ 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 ______

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ equals

$lim_{x \to 0} csc x (2 cos^{2} x + 3 cos x - cos^{2} x + sin x + 4)$ is:

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to π /4} \frac{c o t ^{3} x - t a n x}{c o s ( x + π /4 )}$ is :