Question Number 140905 by bramlexs22 last updated on 14/May/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{x}\:\left(\mathrm{tan}\:\mathrm{x}−\mathrm{x}\right)}\:=? \\ $$ Answered by bramlexs22 last updated on 14/May/21 Answered by mathmax by abdo…
Question Number 9768 by ridwan balatif last updated on 01/Jan/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin3}{x}\mathrm{cos3}{x}−\mathrm{sin3}{x}}{\mathrm{2}{x}^{\mathrm{2}} \mathrm{tan2}{x}\mathrm{cos}\frac{\mathrm{1}}{\mathrm{2}}{x}}=…? \\ $$ Answered by sandy_suhendra last updated on 01/Jan/17 $$=\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\frac{\mathrm{sin3x}\:\left(\mathrm{cos3x}−\mathrm{1}\right)}{\mathrm{2x}^{\mathrm{2}} \:\mathrm{tan2x}\:\mathrm{cos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}…
Question Number 140825 by liberty last updated on 13/May/21 $$\mathrm{Use}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{comparison}\:\mathrm{test} \\ $$$$\mathrm{to}\:\mathrm{determine}\:\mathrm{if}\:\mathrm{the}\:\mathrm{series}\:\mathrm{converges} \\ $$$$\mathrm{or}\:\mathrm{diverges}\: \\ $$$$\:\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{7}+\mathrm{8n}\:\mathrm{ln}\:\left(\mathrm{ln}\:\mathrm{n}\right)}.\: \\ $$ Answered by mathmax by abdo…
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Question Number 75230 by aliesam last updated on 08/Dec/19 Commented by mathmax by abdo last updated on 08/Dec/19 $${e}^{\frac{\mathrm{1}}{{x}}} =\mathrm{1}+\frac{\mathrm{1}}{{x}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:\Rightarrow{x}\:{e}^{\frac{\mathrm{1}}{{x}}} ={x}+\mathrm{1}+{o}\left(\frac{\mathrm{1}}{{x}}\right)\:\:\left({x}\rightarrow+\infty\right) \\ $$$${e}^{\frac{\mathrm{1}}{{x}+\mathrm{1}}} =\mathrm{1}+\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
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Question Number 9654 by ridwan balatif last updated on 22/Dec/16 $$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{ax}+\mathrm{b}−\sqrt{\mathrm{x}}}{\mathrm{x}−\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}},\:\:\mathrm{a}+\mathrm{b}=…? \\ $$ Answered by prakash jain last updated on 22/Dec/16 $$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{ax}+\mathrm{b}−\sqrt{\mathrm{x}}}{\mathrm{x}−\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}},\:\:\mathrm{a}+\mathrm{b}=…? \\…
Question Number 140574 by john_santu last updated on 09/May/21 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{cos}\:\mathrm{2}{x}}\:=? \\ $$ Answered by bramlexs22 last updated on 09/May/21 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\right)}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}}…
Question Number 9488 by Joel575 last updated on 10/Dec/16 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}\:+\:\mathrm{4}\sqrt{\mathrm{2016}}\:+\:\mathrm{9}\sqrt{\mathrm{2016}}\:+\:…\:+\:{x}^{\mathrm{2}} \sqrt{\mathrm{2016}}}{{x}^{\mathrm{3}} } \\ $$ Answered by mrW last updated on 10/Dec/16 $$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}×\left(\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}}…