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Category: Limits

Question-73518

Question Number 73518 by aliesam last updated on 13/Nov/19 Commented by mathmax by abdo last updated on 13/Nov/19 letA(x)=(x+1x1)2x1A(x)=e(2x1)ln(x+1x1)$$={e}^{\left(\mathrm{2}{x}−\mathrm{1}\right){ln}\left(\frac{{x}−\mathrm{1}+\mathrm{2}}{{x}−\mathrm{1}}\right)} \:={e}^{\left(\mathrm{2}{x}−\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{1}}\right)} \:\:{we}\:{have}\:{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{1}}\right)\sim\frac{\mathrm{2}}{{x}−\mathrm{1}}\left({x}\rightarrow+\infty\right) \

lim-x-pi-2-2-1-sin-x-2-cos-2-x-

Question Number 138996 by bramlexs22 last updated on 21/Apr/21 limxπ/221+sinx2cos2x=? Answered by EDWIN88 last updated on 21/Apr/21 $$\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)}.\underset{{x}\rightarrow\pi/\mathrm{2}}…

Question-138959

Question Number 138959 by mathlove last updated on 20/Apr/21 Answered by mathmax by abdo last updated on 20/Apr/21 $$\mathrm{a}^{\mathrm{x}^{\mathrm{2}} −\mathrm{1}} \:=\mathrm{e}^{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{loga}} \:\sim\mathrm{1}+\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{loga}\:\mathrm{and} \