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Question Number 35736 by rahul 19 last updated on 22/May/18

lim_(x→0) ((ln (sec (ex)sec (e^2 x)......sec (e^(50) x)))/(e^2 −e^(2cos x) )) = ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sec}\:\left({ex}\right)\mathrm{sec}\:\left({e}^{\mathrm{2}} {x}\right)......\mathrm{sec}\:\left({e}^{\mathrm{50}} {x}\right)\right)}{{e}^{\mathrm{2}} −{e}^{\mathrm{2cos}\:{x}} }\:=\:? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/May/18

this limit is((0/0)) form =((lim)/(x→0))((lnsec(ex)+lnsec(e^2 x)+...+lnsec(e^(50) x))/(e^2 −e^(2cosx) )) using l,hospital rule =((lim)/(x→0))((tan(ex)+tan(e^2 x)+..+tan(e^(50) x))/(0−e^(2cosx) ×(−2sinx))) deviding N_r and D_r by x =((lim)/(x→0))((e((tan(ex))/(ex))+e^2 ((tan(e^2 x))/(e^2 x))+..+e^(50) ((tan(e^(50) x))/(e^(50) x)))/(2e^(2cosx) ×((sinx)/x))) =((e+e^2 +e^3 +...+e^(50) )/(2e^2 )) =((e(((e^(50) −1)/(e−1))))/(2e^2 )) =(1/2)×(1/e)×((e^(50) −1)/(e−1))

$${this}\:{limit}\:{is}\left(\frac{\mathrm{0}}{\mathrm{0}}\right)\:{form}\:\: \\ $$$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{{lnsec}\left({ex}\right)+{lnsec}\left({e}^{\mathrm{2}} {x}\right)+...+{lnsec}\left({e}^{\mathrm{50}} {x}\right)}{{e}^{\mathrm{2}} −{e}^{\mathrm{2}{cosx}} }\: \\ $$$${using}\:{l},{hospital}\:{rule} \\ $$$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{{tan}\left({ex}\right)+{tan}\left({e}^{\mathrm{2}} {x}\right)+..+{tan}\left({e}^{\mathrm{50}} {x}\right)}{\mathrm{0}−{e}^{\mathrm{2}{cosx}} ×\left(−\mathrm{2}{sinx}\right)} \\ $$$${deviding}\:{N}_{{r}} \:{and}\:{D}_{{r}} \:{by}\:{x} \\ $$$$=\frac{{lim}}{{x}\rightarrow\mathrm{0}}\frac{{e}\frac{{tan}\left({ex}\right)}{{ex}}+{e}^{\mathrm{2}} \frac{{tan}\left({e}^{\mathrm{2}} {x}\right)}{{e}^{\mathrm{2}} {x}}+..+{e}^{\mathrm{50}} \frac{{tan}\left({e}^{\mathrm{50}} {x}\right)}{{e}^{\mathrm{50}} {x}}}{\mathrm{2}{e}^{\mathrm{2}{cosx}} ×\frac{{sinx}}{{x}}} \\ $$$$=\frac{{e}+{e}^{\mathrm{2}} +{e}^{\mathrm{3}} +...+{e}^{\mathrm{50}} }{\mathrm{2}{e}^{\mathrm{2}} } \\ $$$$=\frac{{e}\left(\frac{{e}^{\mathrm{50}} −\mathrm{1}}{{e}−\mathrm{1}}\right)}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{{e}}×\frac{{e}^{\mathrm{50}} −\mathrm{1}}{{e}−\mathrm{1}} \\ $$

Commented by rahul 19 last updated on 25/May/18

Thank You Sir . There is a slight error in 2nd line .�� It will be e*tan(ex) ,......

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