Question Number 85021 by M±th+et£s last updated on 18/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{tan}^{\mathrm{4}} \left({x}\right)\:{cot}\left({ln}^{\mathrm{3}} \left({x}+\mathrm{1}\right)\right){ln}\left({sin}^{\mathrm{3}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)}{{sin}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\:−\sqrt{\mathrm{2}}\right){ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 84932 by jagoll last updated on 17/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{38x}−\mathrm{38sin}\:\mathrm{x}}{\mathrm{19x}^{\mathrm{3}} }\:=\: \\ $$ Commented by john santu last updated on 17/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{38cos}\:\mathrm{38x}\:−\:\mathrm{38cos}\:\mathrm{x}}{\mathrm{3}.\mathrm{19x}^{\mathrm{2}} }\:=…
Question Number 84902 by john santu last updated on 17/Mar/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{5}^{\mathrm{x}} +\mathrm{5}^{\mathrm{2x}} \right)\:^{\frac{\mathrm{1}}{\mathrm{x}}} \:? \\ $$ Commented by john santu last updated on 17/Mar/20…
Question Number 84873 by john santu last updated on 17/Mar/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\left(\frac{\mathrm{x}!}{\mathrm{x}}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$ Commented by john santu last updated on 17/Mar/20 $$\mathrm{yes}.\:\mathrm{i}\:\mathrm{agree}\:\mathrm{sir}…
Question Number 84834 by jagoll last updated on 16/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}\:\mathrm{tan}\:\mathrm{x}}}{\mathrm{sin}\:\mathrm{3x}} \\ $$ Commented by john santu last updated on 16/Mar/20 $$\mathrm{do}\:\mathrm{not}\:\mathrm{exist} \\ $$ Answered…
Question Number 19271 by Joel577 last updated on 08/Aug/17 $$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left(\mathrm{2}\:−\:\mathrm{cos}^{\mathrm{2}} \:{x}\right)^{\frac{\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{cos}\:{x}\right)}}{\left({x}\:−\:\pi\right)^{\mathrm{3}} }} \\ $$ Answered by ajfour last updated on 09/Aug/17 $$=\underset{{x}\rightarrow\pi} {\mathrm{lim}}\left\{\left[\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \left(\pi−\mathrm{x}\right)\right]^{\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}}…
Question Number 84600 by john santu last updated on 14/Mar/20 $$\underset{{x}\rightarrow\infty\:} {\mathrm{lim}}\:{x}\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}\:−\:\frac{{x}}{\mathrm{3}} \\ $$ Commented by jagoll last updated on 14/Mar/20 $$\mathrm{nice}\:\mathrm{mister} \\ $$ Commented…
Question Number 84568 by jagoll last updated on 14/Mar/20 $$\mathrm{without}\:\mathrm{L}'\mathrm{hopital} \\ $$$$\underset{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{2x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\sqrt{\mathrm{a}+\mathrm{bx}}}{\mathrm{4x}^{\mathrm{2}} −\mathrm{1}}\:=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{b} \\ $$ Commented by john santu last…
Question Number 84492 by john santu last updated on 13/Mar/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)}{\mathrm{1}−\mathrm{2cos}\:\left({x}\right)}\:=\: \\ $$ Commented by john santu last updated on 13/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}\right)}{\mathrm{1}−\mathrm{2cos}\:\left({x}+\frac{\pi}{\mathrm{3}}\right)}\:= \\…
Question Number 84460 by jagoll last updated on 13/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{2}+\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{2}−\mathrm{x}\right)}{\mathrm{x}} \\ $$ Commented by jagoll last updated on 13/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\mathrm{2}+\mathrm{x}\right)+\mathrm{cos}\:\left(\mathrm{2}−\mathrm{x}\right)}{\mathrm{1}} \\ $$$$=\:\mathrm{2cos}\:\mathrm{2} \\…