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Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 28 January, Morning Shift — Question 13

The value of lim⁡x→0log⁡e(sec⁡(ex)⋅sec⁡(e2x)⋅…⋅sec⁡(e10x))e2−e2cos⁡x\lim _{x \rightarrow 0} \frac{\log _{e}\left(\sec (e x) \cdot \sec \left(e^{2} x\right) \cdot \ldots \cdot \sec \left(e^{10} x\right)\right)}{e^{2}-e^{2 \cos x}} is equal to

  1. Option A:

    (e10−1)2e2(e2−1)\frac{\left(\mathrm{e}^{10}-1\right)}{2 \mathrm{e}^{2}\left(\mathrm{e}^{2}-1\right)}

  2. Option B:

    (e20−1)2e2(e2−1)\frac{\left(\mathrm{e}^{20}-1\right)}{2 \mathrm{e}^{2}\left(\mathrm{e}^{2}-1\right)}

  3. Option C:

    (e20−1)2(e2−1)\frac{\left(\mathrm{e}^{20}-1\right)}{2\left(\mathrm{e}^{2}-1\right)}

    Correct
  4. Option D:

    (e10−1)2(e2−1)\frac{\left(\mathrm{e}^{10}-1\right)}{2\left(\mathrm{e}^{2}-1\right)}

Answer: C

Step-by-step solution

⇒lim⁡x→0ln⁡(sec⁡(ex))+ln⁡(sec⁡(e2x))+…..ℓn(sec⁡(e10x))e2cos⁡x(e2−2cos⁡x−12−2cos⁡x)×2−2cos⁡xx2×x2\Rightarrow \lim _{\mathrm{x} \rightarrow 0} \frac{\ln (\sec (\mathrm{ex}))+\ln \left(\sec \left(\mathrm{e}^{2} \mathrm{x}\right)\right)+\ldots . . \ell \mathrm{n}\left(\sec \left(\mathrm{e}^{10} \mathrm{x}\right)\right)}{\mathrm{e}^{2 \cos \mathrm{x}}\left(\frac{\mathrm{e}^{2-2 \cos \mathrm{x}}-1}{2-2 \cos \mathrm{x}}\right) \times \frac{2-2 \cos \mathrm{x}}{\mathrm{x}^{2}} \times \mathrm{x}^{2}}

⇒lim⁡x→10ℓn(sec⁡(ex))+ℓn(sec⁡(e2x))+……ℓn(sec⁡(e10x))e2x2\Rightarrow \lim _{\mathrm{x} \rightarrow 10} \frac{\ell \mathrm{n}(\sec (\mathrm{ex}))+\ell \mathrm{n}\left(\sec \left(\mathrm{e}^{2} \mathrm{x}\right)\right)+\ldots \ldots \ell \mathrm{n}\left(\sec \left(\mathrm{e}^{10} \mathrm{x}\right)\right)}{\mathrm{e}^{2} \mathrm{x}^{2}}

Using L'H rule ⇒lim⁡x→10etan⁡ex+e2tan⁡2x+…..+e10tan⁡10x2e2x\Rightarrow \lim _{\mathrm{x} \rightarrow 10} \frac{\mathrm{e} \tan \mathrm{ex}+\mathrm{e}^{2} \tan ^{2} \mathrm{x}+\ldots . .+\mathrm{e}^{10} \tan ^{10} \mathrm{x}}{2 \mathrm{e}^{2} \mathrm{x}}

⇒12e2[e2+e4+e6+….+e20]\Rightarrow \frac{1}{2 \mathrm{e}^{2}}\left[\mathrm{e}^{2}+\mathrm{e}^{4}+\mathrm{e}^{6}+\ldots .+\mathrm{e}^{20}\right]

⇒12e2((e2)10−1)e2(e2−1)\Rightarrow \frac{1}{2} \frac{\mathrm{e}^{2}\left(\left(\mathrm{e}^{2}\right)^{10}-1\right)}{\mathrm{e}^{2}\left(\mathrm{e}^{2}-1\right)}

⇒12(e20−1)(e2−1)\Rightarrow \frac{1}{2} \frac{\left(\mathrm{e}^{20}-1\right)}{\left(\mathrm{e}^{2}-1\right)}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Morning Shift
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Evaluation of Limit of Functions