The value of $\lim_{x \to 0} \frac{\log_e(\sec(e^x) \cdot \sec(e^{2x}) \cdot \dots \cdot \sec(e^{10x}))}{e^2 - e^{2 \cos x}}$ is equal to Options: (A) $\frac{(e^{20}-1)}{2(e^2-1)}$ (B) $\frac{(e^{10}-1)}{2(e^2-1)}$ (C) $\frac{(e^{10}-1)}{2e^2(e^2-1)}$ (D) $\frac{(e^{20}-1)}{2e^2(e^2-1)}$

### Analyzing the Setup The given limit involves a product of ten secant terms nested within a logarithm, divided by a difference of exponentials. To simplify the numerator, we utilize the logarithmic identity $\log(abc) = \log a + \log b + \log c$. The numerator expression becomes: $$\sum_{k=1}^{10} \log_e(\sec(x e^k))$$ ### Simplifying the Numerator As $x \to 0$, the argument $\theta = x e^k$ also approaches zero. We utilize the Maclaurin series expansion for $\sec \theta \approx 1 + \frac{\theta^2}{2}$. Substituting this into the natural logarithm, we apply the approximation $\ln(1+u) \approx u$ for small $u$: $$\log_e\left(1 + \frac{\theta^2}{2}\right) \approx \frac{\theta^2}{2} = \frac{(x e^k)^2}{2} = \frac{x^2 e^{2k}}{2}$$ Summing these terms, we factor out the constant $\frac{x^2}{2}$: $$\frac{x^2}{2} \sum_{k=1}^{10} e^{2k}$$ The summation is a **Geometric Progression** with $a = e^2$, $r = e^2$, and $n = 10$. Using the sum formula $S = a \frac{r^n - 1}{r - 1}$, the numerator simplifies to: $$\frac{x^2}{2} \cdot e^2 \left( \frac{(e^2)^{10} - 1}{e^2 - 1} \right) = \frac{x^2 e^2 (e^{20} - 1)}{2(e^2 - 1)}$$ ### Simplifying the Denominator The denominator is $e^2 - e^{2 \cos x}$. We factor out $e^2$ to obtain $e^2(1 - e^{2(\cos x - 1)})$. Using the small-angle approximation $\cos x \approx 1 - \frac{x^2}{2}$, the exponent becomes $2(\cos x - 1) \approx -x^2$. Applying the expansion $e^u \approx 1 + u$: $$e^2(1 - e^{-x^2}) \approx e^2(1 - (1 - x^2)) = e^2 x^2$$ ### Final Calculation Now, we combine the simplified numerator and denominator to evaluate the limit as $x \to 0$: $$\lim_{x \to 0} \frac{\frac{x^2 e^2 (e^{20} - 1)}{2(e^2 - 1)}}{e^2 x^2}$$ The $x^2$ and $e^2$ terms cancel out, leaving the final result: $$\mathbf{\frac{e^{20} - 1}{2(e^2 - 1)}}$$

How did you get $\ln(\sec \theta) \approx \frac{\theta^2}{2}$ so quickly?

This comes from the Maclaurin series. We know $\sec \theta = 1 + \frac{\theta^2}{2} + \frac{5\theta^4}{24} + \dots$. When we take the natural log, $\ln(1 + \frac{\theta^2}{2} + \dots)$, we use the expansion $\ln(1+u) \approx u$. Thus, $\ln(1 + \frac{\theta^2}{2}) \approx \frac{\theta^2}{2}$. It is a standard result you should memorize for competitive exams to save precious time.

JEE Main 2026 (28 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of $lim_{x \to 0} \frac{l o g _{e} ( s e c ( e ^{x} ) \cdot s e c ( e ^{2 x} ) \cdot \dots \cdot s e c ( e ^{10 x} ))}{e ^{2} - e ^{2 c o s x}}$ is equal to

Q: Why did we factor out $e^2$ in the denominator? Why not just expand $e^{2 \cos x}$ directly?

If you expand $e^{2 \cos x}$ directly, you get $e^{2(1 - x^2/2)} = e^{2 - x^2} = e^2 \cdot e^{-x^2}$. This is exactly what we did by factoring! Factoring $e^2$ first makes the term $e^{-x^2}$ visible, which allows us to use the $e^u \approx 1+u$ approximation easily. It keeps the algebra clean and prevents sign errors.

Select Answer:

Progressive Hints

Hint 1

Do not let the product of ten secant terms intimidate you. The first step is to use the fundamental property of logarithms: the logarithm of a product is the sum of the logarithms. This transforms a terrifying, complex product into a manageable summation that we can handle term by term.

Hint 2

For small angles, we need to approximate the behavior of the function

ln (sec θ)

. Recall the Maclaurin series expansion:

sec θ \approx 1 + \frac{θ ^{2}}{2}

and

ln (1 + t) \approx t

for small

t

. Combine these to find that

ln (sec θ) \approx \frac{θ ^{2}}{2}

. This is the key to unlocking the numerator.

Hint 3

In the denominator, the expression

e^{2} - e^{2 c o s x}

is a classic indeterminate form. Factor out

e^{2}

to get

e^{2} (1 - e^{2 (c o s x - 1)})

. Since

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

, the exponent becomes

- x^{2}

. Use the approximation

e^{u} \approx 1 + u

to simplify the denominator into a simple

e^{2} x^{2}

term, which will beautifully cancel out the

x^{2}

from the numerator.

Visualized Solution

Analyze the Limit Structure

Given limit: $lim_{x \to 0} \frac{l o g _{e} ( s e c ( x e ^{1} ) \cdot s e c ( x e ^{2} ) \dots s e c ( x e ^{10} ))}{e ^{2} - e ^{2 c o s x}}$
As $x \to 0$ , the numerator approaches $lo g_{e} (1) = 0$ .
The denominator approaches $e^{2} - e^{2} = 0$ .
This is a $\frac{0}{0}$ indeterminate form.
We will use Taylor Series expansions to simplify the terms.

Simplify the Numerator

Let the numerator be $L_{n u m} = lo g_{e} (\prod_{k = 1}^{10} sec (x e^{k}))$
Using the property $lo g (a \cdot b \cdot c) = lo g a + lo g b + lo g c$ :
$L_{n u m} = \sum_{k = 1}^{10} lo g_{e} (sec (x e^{k}))$

Expansion of $lo g (sec θ)$

For a small angle $θ$ , the Maclaurin series gives:
$sec θ \approx 1 + \frac{θ ^{2}}{2}$
We also know that $lo g_{e} (1 + t) \approx t$ for small $t$ .
Therefore, $lo g_{e} (sec θ) \approx lo g_{e} (1 + \frac{θ ^{2}}{2}) \approx \frac{θ ^{2}}{2}$

Apply Expansion to the Numerator

Substitute $θ = x e^{k}$ into our approximation:
$lo g_{e} (sec (x e^{k})) \approx \frac{( x e ^{k} ) ^{2}}{2} = \frac{x ^{2} e ^{2 k}}{2}$
Summing over all terms from $k = 1$ to $10$ :
$L_{n u m} \approx \sum_{k = 1}^{10} \frac{x ^{2} e ^{2 k}}{2} = \frac{x ^{2}}{2} \sum_{k = 1}^{10} e^{2 k}$

Summing the Geometric Progression

The series $\sum_{k = 1}^{10} e^{2 k} = e^{2} + e^{4} + \dots + e^{20}$ is a Geometric Progression.
First term $a = e^{2}$ , common ratio $r = e^{2}$ , number of terms $n = 10$ .
Sum $S_{10} = a \frac{r ^{n} - 1}{r - 1} = e^{2} \frac{( e ^{2} ) ^{10} - 1}{e ^{2} - 1} = \frac{e ^{2} ( e ^{20} - 1 )}{e ^{2} - 1}$
Numerator $\approx \frac{x ^{2} e ^{2} ( e ^{20} - 1 )}{2 ( e ^{2} - 1 )}$

Analyze the Denominator

Now consider the denominator: $D = e^{2} - e^{2 c o s x}$
Factor out $e^{2}$ to simplify the expression:
$D = e^{2} (1 - e^{2 c o s x - 2})$
$D = e^{2} (1 - e^{2 (c o s x - 1)})$

Expand the Denominator Terms

As $x \to 0$ , we use the expansion $cos x \approx 1 - \frac{x ^{2}}{2}$ .
So, $cos x - 1 \approx - \frac{x ^{2}}{2}$ .
The exponent becomes: $2 (cos x - 1) \approx 2 (- \frac{x ^{2}}{2}) = - x^{2}$ .
Using $e^{t} \approx 1 + t$ for small $t$ :
$D \approx e^{2} (1 - (1 - x^{2})) = e^{2} x^{2}$

Combine and Evaluate the Limit

Substitute the simplified numerator and denominator back into the limit:
$lim_{x \to 0} \frac{\frac{x ^{2} e ^{2} ( e ^{20} - 1 )}{2 ( e ^{2} - 1 )}}{e ^{2} x ^{2}}$
Cancel out the common terms $x^{2}$ and $e^{2}$ :
Limit = $\frac{e ^{20} - 1}{2 ( e ^{2} - 1 )}$

Final Conclusion

Final Result: $\frac{e ^{20} - 1}{2 ( e ^{2} - 1 )}$
Key Takeaways:
1. Convert the log of a product into a sum of logs.
2. Use the approximation $lo g (sec θ) \approx \frac{θ ^{2}}{2}$ for small $θ$ .
3. Use $cos x \approx 1 - \frac{x ^{2}}{2}$ and $e^{t} \approx 1 + t$ to simplify the denominator.

00:00 / 00:00

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

Analyzing the Setup

The given limit involves a product of ten secant terms nested within a logarithm, divided by a difference of exponentials. To simplify the numerator, we utilize the logarithmic identity

lo g (ab c) = lo g a + lo g b + lo g c

The numerator expression becomes:

k = 1 \sum 10 lo g_{e} (sec (x e^{k}))

Simplifying the Numerator

x \to 0

, the argument

θ = x e^{k}

also approaches zero. We utilize the Maclaurin series expansion for

sec θ \approx 1 + \frac{θ ^{2}}{2}

Substituting this into the natural logarithm, we apply the approximation

ln (1 + u) \approx u

for small

u

lo g_{e} (1 + \frac{θ ^{2}}{2}) \approx \frac{θ ^{2}}{2} = \frac{( x e ^{k} ) ^{2}}{2} = \frac{x ^{2} e ^{2 k}}{2}

Summing these terms, we factor out the constant

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

\frac{x ^{2}}{2} k = 1 \sum 10 e^{2 k}

The summation is a Geometric Progression with

a = e^{2}

r = e^{2}

, and

n = 10

. Using the sum formula

S = a \frac{r ^{n} - 1}{r - 1}

, the numerator simplifies to:

\frac{x ^{2}}{2} \cdot e^{2} (\frac{( e ^{2} ) ^{10} - 1}{e ^{2} - 1}) = \frac{x ^{2} e ^{2} ( e ^{20} - 1 )}{2 ( e ^{2} - 1 )}

Simplifying the Denominator

The denominator is

e^{2} - e^{2 c o s x}

. We factor out

e^{2}

to obtain

e^{2} (1 - e^{2 (c o s x - 1)})

Using the small-angle approximation

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

, the exponent becomes

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

. Applying the expansion

e^{u} \approx 1 + u

e^{2} (1 - e^{- x^{2}}) \approx e^{2} (1 - (1 - x^{2})) = e^{2} x^{2}

Final Calculation

Now, we combine the simplified numerator and denominator to evaluate the limit as

x \to 0

x \to 0 lim \frac{\frac{x ^{2} e ^{2} ( e ^{20} - 1 )}{2 ( e ^{2} - 1 )}}{e ^{2} x ^{2}}

The

x^{2}

and

e^{2}

terms cancel out, leaving the final result:

\frac{e ^{20} - 1}{2 ( e ^{2} - 1 )}

Common Traps & FAQs

Why can't we just use L'Hopital's Rule here? It seems like a standard

0/0

form.

While L'Hopital's Rule is mathematically valid, it is a trap! Differentiating a product of ten secant functions using the product rule would be a nightmare of algebraic complexity. In JEE Advanced, we look for elegance, not brute force. Taylor series expansions are the scalpel that cuts through this complexity cleanly and efficiently.

How did you get

ln (sec θ) \approx \frac{θ ^{2}}{2}

so quickly?

This comes from the Maclaurin series. We know

sec θ = 1 + \frac{θ ^{2}}{2} + \frac{5 θ ^{4}}{24} + \dots

. When we take the natural log,

ln (1 + \frac{θ ^{2}}{2} + \dots)

, we use the expansion

ln (1 + u) \approx u

. Thus,

ln (1 + \frac{θ ^{2}}{2}) \approx \frac{θ ^{2}}{2}

. It is a standard result you should memorize for competitive exams to save precious time.

Why did we factor out

e^{2}

in the denominator? Why not just expand

e^{2 c o s x}

directly?

If you expand

e^{2 c o s x}

directly, you get

e^{2 (1 - x^{2} /2)} = e^{2 - x^{2}} = e^{2} \cdot e^{- x^{2}}

. This is exactly what we did by factoring! Factoring

e^{2}

first makes the term

e^{- x^{2}}

visible, which allows us to use the

e^{u} \approx 1 + u

approximation easily. It keeps the algebra clean and prevents sign errors.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of $lim_{x \to 0} \frac{l o g _{e} ( s e c ( e ^{x} ) \cdot s e c ( e ^{2 x} ) \cdot \dots \cdot s e c ( e ^{10 x} ))}{e ^{2} - e ^{2 c o s x}}$ is equal to

Select Answer:

Visualized Solution

Analyze the Limit Structure

Simplify the Numerator

Expansion of $lo g (sec θ)$

Apply Expansion to the Numerator

Summing the Geometric Progression

Analyze the Denominator

Expand the Denominator Terms

Combine and Evaluate the Limit

Final Conclusion

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

Analyzing the Setup

Simplifying the Numerator

Simplifying the Denominator

Final Calculation

Similar Questions

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

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

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

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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

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If $lim_{x \to \infty} ((\frac{e}{1 - e}) (\frac{1}{e} - \frac{x}{1 + x}))^{x} = α$ , then the value of $\frac{l o g _{e} α}{1 + l o g _{e} α}$ equals

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​e2−e2cosxloge​(sec(ex)⋅sec(e2x)⋅⋯⋅sec(e10x))​ is equal to

Select Answer:

Visualized Solution

Analyze the Limit Structure

Simplify the Numerator

Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } log(secθ)

Apply Expansion to the Numerator

Summing the Geometric Progression

Analyze the Denominator

Expand the Denominator Terms

Combine and Evaluate the Limit

Final Conclusion

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

Analyzing the Setup

Simplifying the Numerator

Simplifying the Denominator

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of $lim_{x \to 0} \frac{l o g _{e} ( s e c ( e ^{x} ) \cdot s e c ( e ^{2 x} ) \cdot \dots \cdot s e c ( e ^{10 x} ))}{e ^{2} - e ^{2 c o s x}}$ is equal to

Expansion of $lo g (sec θ)$

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

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

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

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

$lim_{x \to 0} (\frac{( x + 2 c o s x ) ^{3} + 2 ( x + 2 c o s x ) ^{2} + 3 s i n ( x + 2 c o s x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 s i n ( x + 2 )})^{\frac{100}{x}}$ is equal to ______.

$lim_{x \to 0} (tan (\frac{π}{4} + x))^{\frac{1}{x}}$ is equal to :

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

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

If $lim_{x \to \infty} ((\frac{e}{1 - e}) (\frac{1}{e} - \frac{x}{1 + x}))^{x} = α$ , then the value of $\frac{l o g _{e} α}{1 + l o g _{e} α}$ equals