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Category: Limits

Question-72251

Question Number 72251 by aliesam last updated on 26/Oct/19 Commented by mathmax by abdo last updated on 26/Oct/19 letf(x)=ln(1+ax+bx2)ln(1ax+bx2)1cosx$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{o}\left({u}\right)\:\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow \

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

Question Number 137773 by Ajadiazeemolamilekan last updated on 06/Apr/21 limx0cospxcosqxx2 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}} } \