lim_{x to 0} frac{log_{e}(sec(ex) cdot sec(e^ | vedclass

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(C) ધારો કે $L = \lim_{x 	o 0} \frac{\ln(\sec(ex)) + \ln(\sec(e^{2}x)) + ... + \ln(\sec(e^{10}x))}{e^{2} - e^{2\cos x}}$.$\ln(\sec 	heta) \approx \f

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

Vedclass · Answer

(C) ધારો કે $L = \lim_{x 	o 0} \frac{\ln(\sec(ex)) + \ln(\sec(e^{2}x)) + ... + \ln(\sec(e^{10}x))}{e^{2} - e^{2\cos x}}$.$\ln(\sec 	heta) \approx \frac{	heta^{2}}{2}$ નો ઉપયોગ કરતા, અંશ $\frac{x^{2}}{2} \sum_{k=1}^{10} e^{2k}$ થશે.છેદ $e^{2} - e^{2\cos x} = e^{2}(1 - e^{2\cos x - 2}) \approx 2e^{2}(1 - \cos x) \approx e^{2}x^{2}$ થશે.તેથી, $L = \lim_{x 	o 0} \frac{\frac{x^{2}}{2} \sum_{k=1}^{10} e^{2k}}{e^{2}x^{2}} = \frac{1}{2e^{2}} \sum_{k=1}^{10} (e^{2})^{k}$.ભૌમિતિક શ્રેણીના સરવાળાના સૂત્ર મુજબ, $L = \frac{1}{2e^{2}} \cdot \frac{e^{2}(e^{20} - 1)}{e^{2} - 1} = \frac{e^{20} - 1}{2(e^{2} - 1)}$.

$\lim_{x \to 0} \frac{\log_{e}(\sec(ex) \cdot \sec(e^{2}x) \cdot ... \cdot \sec(e^{10}x))}{e^{2} - e^{2\cos x}}$ ની કિંમત શોધો.

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