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lim-x-0-cos-px-cos-qx-x-2-

Question Number 137773 by Ajadiazeemolamilekan last updated on 06/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{px}−\mathrm{cos}\:{qx}}{{x}^{\mathrm{2}} } \\ $$ Answered by bemath last updated on 06/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:\left(\frac{{p}+{q}}{\mathrm{2}}\:{x}\right)\mathrm{sin}\:\left(\frac{{p}−{q}}{\mathrm{2}}\:{x}\right)}{{x}^{\mathrm{2}} } \\…

l-x-0-im-x-sin-x-x-2-e-x-1-

Question Number 6703 by love math last updated on 15/Jul/16 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{l}im}\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \left({e}^{{x}} −\mathrm{1}\right)} \\ $$ Answered by FilupSmith last updated on 15/Jul/16 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}}…

Question-137772

Question Number 137772 by peter frank last updated on 06/Apr/21 Answered by bemath last updated on 06/Apr/21 $$\Leftrightarrow\:\frac{\left(\mathrm{2}^{{n}} \right)^{\mathrm{3}} +\left(\mathrm{3}^{{n}} \right)^{\mathrm{3}} }{\left(\mathrm{2}^{{n}} \right)^{\mathrm{2}} \left(\mathrm{3}\right)^{{n}} +\mathrm{2}^{{n}}…

to-all-those-who-deleted-their-posts-after-they-had-been-answered-I-will-not-answer-you-anymore-this-forum-had-been-great-but-lately-it-has-been-filling-with-unpolite-people-I-m-not-a-freebie-solver-

Question Number 72232 by MJS last updated on 26/Oct/19 $$\mathrm{to}\:\mathrm{all}\:\mathrm{those}\:\mathrm{who}\:\mathrm{deleted}\:\mathrm{their}\:\mathrm{posts}\:\mathrm{after}\:\mathrm{they} \\ $$$$\mathrm{had}\:\mathrm{been}\:\mathrm{answered}:\:\mathrm{I}\:\mathrm{will}\:\mathrm{not}\:\mathrm{answer}\:\mathrm{you} \\ $$$$\mathrm{anymore} \\ $$$$\mathrm{this}\:\mathrm{forum}\:\mathrm{had}\:\mathrm{been}\:\mathrm{great}\:\mathrm{but}\:\mathrm{lately}\:\mathrm{it}\:\mathrm{has} \\ $$$$\mathrm{been}\:\mathrm{filling}\:\mathrm{with}\:\mathrm{unpolite}\:\mathrm{people} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{a}\:\mathrm{freebie}\:\mathrm{solver}\:\mathrm{for}\:\mathrm{anybody}'\mathrm{s}\:\mathrm{homework} \\ $$ Commented by mr…

Question-137770

Question Number 137770 by peter frank last updated on 06/Apr/21 Answered by Ñï= last updated on 06/Apr/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {da}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\left(\mathrm{1}+{ax}\right)\left(\mathrm{1}+{x}^{\mathrm{2}}…