The value of lim(x→0) 2 * (1 - cos x * √(cos 2x) * ∛(cos 3x) * ... * ¹⁰√(cos 10x)) / x² is

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

While L'Hopital's rule is a valid tool, applying it to a product of ten different radical terms would lead to a nightmare of differentiation using the product rule and chain rule. It is far more elegant and less error-prone to use Taylor series expansions to linearize the expression near $x = 0$.

Q: Why does the product $\prod_{k=1}^{10} (1 - a_k)$ simplify to $1 - \sum_{k=1}^{10} a_k$?

When you expand the product $(1 - a_1)(1 - a_2)\dots(1 - a_{10})$, you get $1 - (a_1 + a_2 + \dots + a_{10}) + (a_1 a_2 + a_1 a_3 + \dots) - \dots$. Since each $a_k$ contains an $x^2$ term, the products of two or more $a_k$ terms will contain $x^4, x^6$, and higher powers. As $x \to 0$, these higher-order terms vanish much faster than the $x^2$ terms, so we can safely ignore them.

Q: Is it mathematically rigorous to use the binomial approximation $(1 - y)^n \approx 1 - ny$ when $n = 1/k$?

Absolutely! The binomial expansion $(1 - y)^n = 1 - ny + \frac{n(n-1)}{2}y^2 - \dots$ is valid for any real exponent $n$ when $|y| < 1$. Since $y = \frac{k^2 x^2}{2}$ approaches $0$ as $x \to 0$, the higher-order terms involving $y^2$ (which would be proportional to $x^4$) become negligible, leaving us with the first-order approximation $1 - ny$.

Question

The value of $\lim_{x 	o 0} 2 \left( \frac{1 - \cos x \sqrt{\cos 2x} \sqrt[3]{\cos 3x} \dots \sqrt[10]{\cos 10x}}{x^2} ight)$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a towering, jagged mountain of a limit problem:
$$\lim_{x 	o 0} 2 \left( \frac{1 - \prod_{k=1}^{10} \sqrt[k]{\cos kx}}{x^2} ight)$$

At first glance, it looks impossible. However, as a student of the JEE, you know that every monster has a weakness. Our weakness here is the fact that $x$ is approaching zero, where functions behave in predictable, linear ways.

### Phase 1: The General Term Strategy

The numerator is a product of ten different terms, each with a different root. Trying to differentiate this directly is a trap. Instead, let us isolate the general term, $T_k = \sqrt[k]{\cos kx}$.

By rewriting this with fractional exponents, we get $T_k = (\cos kx)^{\frac{1}{k}}$. Now, we use the Maclaurin series expansion. We know that for small $	heta$, $\cos 	heta \approx 1 - \frac{	heta^2}{2}$.

Substituting $	heta = kx$, we find:
$$\cos kx \approx 1 - \frac{k^2 x^2}{2}$$

### Phase 2: The Taylor-Binomial Dance

Now, substitute this back into our general term:
$$T_k \approx \left( 1 - \frac{k^2 x^2}{2} ight)^{\frac{1}{k}}$$

This is where the magic happens. We have a term of the form $(1 - y)^n$, where $y = \frac{k^2 x^2}{2}$ and $n = \frac{1}{k}$. Since $x$ is small, $y$ is also small.

The binomial approximation $(1 - y)^n \approx 1 - ny$ yields:
$$T_k \approx 1 - \frac{1}{k} \cdot \frac{k^2 x^2}{2} = 1 - \frac{k x^2}{2}$$

### Phase 3: The Product Collapse

We have successfully linearized the general term. Now, the entire product $P = \prod_{k=1}^{10} T_k$ becomes a product of terms like $(1 - \frac{k x^2}{2})$.

Using the property that $\prod (1 - a_k) \approx 1 - \sum a_k$ for small $a_k$, we can write:
$$P \approx 1 - \sum_{k=1}^{10} \frac{k x^2}{2}$$

Factoring out the common $\frac{x^2}{2}$, we get:
$$P \approx 1 - \frac{x^2}{2} \sum_{k=1}^{10} k$$
The monstrous product has collapsed into a simple arithmetic sum.

### Phase 4: The Final Victory

Now, we return to our original limit:
$$L = \lim_{x 	o 0} 2 \left( \frac{1 - P}{x^2} ight)$$

Substituting our simplified $P$, the numerator becomes $1 - (1 - \frac{x^2}{2} \sum_{k=1}^{10} k)$, which simplifies to $\frac{x^2}{2} \sum_{k=1}^{10} k$. The $1$s cancel and the signs flip, leaving us with:
$$L = \lim_{x 	o 0} 2 \cdot \frac{\frac{x^2}{2} \sum_{k=1}^{10} k}{x^2}$$

The $x^2$ terms cancel out, the $2$ and $\frac{1}{2}$ cancel out, and we are left with just the sum of the first ten natural numbers:
$$\sum_{k=1}^{10} k = \frac{10(11)}{2} = 55$$

We have conquered the mountain. The final answer is **55**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Enter Numerical Value:

Visualized Solution

Identify the Limit Form

Analyze the General Term

Apply Taylor Expansion

Apply Binomial Approximation

Simplify the General Term

Expand the Product

Factor out Common Terms

Substitute back into Limit

Simplify the Expression

Final Calculation

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

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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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; } The value of limx→0​2(x21−cosxcos2x​3cos3x​…10cos10x​​) is