Limits Logarithm and Sec Function Trick

 

❓ Question

Evaluate:

$$\lim_{x\to0} \frac{ \ln\Big(\sec(ex)\sec(e^2x)\sec(e^3x)\cdots\sec(e^{10}x)\Big) } {e^2-e^2\cos x}$$

Options:

1. $\dfrac{e^{18}-1}{e^2-1}$
2. $\dfrac{e^{20}-1}{e^2-1}$
3. $\dfrac{e^{16}-1}{e^2-1}$
4. $\dfrac{e^{22}-1}{e^2-1}$

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Standard limits
👉 Logarithmic expansion
👉 Small angle approximation.

Main idea:

Use:

$$\ln(\sec t)\sim \frac{t^2}{2} \quad (t\to0)$$

and:

$$1-\cos x\sim \frac{x^2}{2}$$

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🔷 Step 1 — Convert Product into Sum 💯

Using:

$$\ln(ab)=\ln a+\ln b$$

Numerator becomes:

$$\sum_{k=1}^{10}\ln(\sec(e^k x))$$

So limit:

$$L= \lim_{x\to0} \frac{ \sum_{k=1}^{10}\ln(\sec(e^k x)) } { e^2(1-\cos x) }$$

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🔷 Step 2 — Use Standard Expansion

For small $t$:

$$\sec t = 1+\frac{t^2}{2}+o(t^2)$$

Hence:

$$\ln(\sec t)\sim\frac{t^2}{2}$$

Put:

$$t=e^k x$$

Thus:

$$\ln(\sec(e^k x)) \sim \frac{e^{2k}x^2}{2}$$

Therefore numerator:

$$\sim \frac{x^2}{2} \sum_{k=1}^{10}e^{2k}$$

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🔷 Step 3 — Simplify Denominator

Using:

$$1-\cos x\sim\frac{x^2}{2}$$

Denominator:

$$e^2(1-\cos x) \sim e^2\cdot\frac{x^2}{2}$$

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🔷 Step 4 — Compute Limit

Thus:

$$L= \frac{ \frac{x^2}{2}\sum_{k=1}^{10}e^{2k} } { e^2\frac{x^2}{2} }$$
$$= \frac1{e^2}\sum_{k=1}^{10}e^{2k}$$

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🔷 Step 5 — Evaluate GP

$$\sum_{k=1}^{10}e^{2k} = e^2+e^4+\cdots+e^{20}$$

This is GP with:

$$a=e^2,\quad r=e^2$$

So:

$$= e^2\cdot\frac{e^{20}-1}{e^2-1}$$

Divide by $e^2$:

$$L= \frac{e^{20}-1}{e^2-1}$$

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🔷 JEE Trap Alert 🚨

❌ Directly expanding sec repeatedly

❌ Forgetting:

$$\boxed{ \ln(\sec t)\sim \frac{t^2}{2} }$$

This shortcut saves huge time.

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✅ Final Answer

$$\boxed{ \frac{e^{20}-1}{e^2-1} }$$

(Option 2)

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