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Question-140369




Question Number 140369 by meetbhavsar25 last updated on 06/May/21
Commented by meetbhavsar25 last updated on 08/May/21
mr W, please help me out
$${mr}\:{W},\:{please}\:{help}\:{me}\:{out} \\ $$
Answered by john_santu last updated on 07/May/21
lim_(x→0)  ((e^x −1)/(sin ((x/k))ln (1+(x/4)))) = 12  you can use Taylor series
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −\mathrm{1}}{\mathrm{sin}\:\left(\frac{{x}}{{k}}\right)\mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{\mathrm{4}}\right)}\:=\:\mathrm{12} \\ $$$${you}\:{can}\:{use}\:{Taylor}\:{series} \\ $$
Commented by meetbhavsar25 last updated on 07/May/21
lim_(x→0)  ((e^x −1)/(sin ((x/k))ln (1+(x/4)))) = 12  you can use Taylor series  How??
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −\mathrm{1}}{\mathrm{sin}\:\left(\frac{{x}}{{k}}\right)\mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{\mathrm{4}}\right)}\:=\:\mathrm{12} \\ $$$${you}\:{can}\:{use}\:{Taylor}\:{series} \\ $$$${How}?? \\ $$

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