JEE Mains · Maths · STD 11 - 12. limits
The value of \(lim_{x\rightarrow0}\frac{log_{e}(sec(ex) \cdot sec(e^{2}x)\cdot...\cdot sec(e^{10}x))}{e^{2}-e^{2cos~x}}\) is equal to
- A \(\frac{(e^{10}-1)}{2e^{2}(e^{2}-1)}\)
- B \(\frac{(e^{20}-1)}{2e^{2}(e^{2}-1)}\)
- C \(\frac{(e^{20}-1)}{2(e^{2}-1)}\)
- D \(\frac{(e^{10}-1)}{2(e^{2}-1)}\)
Answer & Solution
Correct Answer
(C) \(\frac{(e^{20}-1)}{2(e^{2}-1)}\)
Step-by-step Solution
Detailed explanation
\(\Rightarrow lim_{x\rightarrow0}\frac{ln(sec(ex))+ln(sec(e^{2}x))+~...~ln(sec(e^{10}x))}{e^{2cosx}(\frac{e^{2-2~cosx}-1}{2-2~cos~x})\times\frac{2-2~cos~x}{x^{2}}\times x^{2}}\)…
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