Question Number 18803 by daffa22 last updated on 30/Jul/17 Commented by daffa22 last updated on 30/Jul/17 $$\mathrm{help} \\ $$ Answered by behi.8.3.4.1.7@gmail.com last updated on…
Question Number 149871 by ArielVyny last updated on 07/Aug/21 $${lim}_{{x}\rightarrow\mathrm{2}} \frac{\mathrm{3}^{{x}!} −\mathrm{9}}{{x}−\mathrm{2}} \\ $$ Answered by Ar Brandon last updated on 08/Aug/21 $$\mathscr{L}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{3}^{{x}!} −\mathrm{9}}{{x}−\mathrm{2}}…
Question Number 84330 by M±th+et£s last updated on 11/Mar/20 $$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\frac{\left(\mathrm{2}{x}^{\mathrm{3}} +{x}+\mathrm{1}\right)−\mathrm{64}}{{x}^{\mathrm{3}} −\mathrm{1}} \\ $$ Commented by M±th+et£s last updated on 11/Mar/20 $${typo}\:\left(\mathrm{2}{x}^{\mathrm{3}} +{x}+\mathrm{1}\right)^{\mathrm{3}} \\…
Question Number 84309 by jagoll last updated on 11/Mar/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{ln}\:\mid\mathrm{x}\mid\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}−\mathrm{ln}\mid\mathrm{x}\mid} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84283 by jagoll last updated on 11/Mar/20 $$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mid\mathrm{x}\mid−\mathrm{1}\right)^{\mathrm{2}} −\left(\mid\mathrm{a}\mid−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}−\mathrm{a}}\:=\:\mathrm{k}\:,\:\mathrm{a}>\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mid\mathrm{x}\mid−\mathrm{1}\right)^{\mathrm{3}} −\left(\mid\mathrm{a}\mid−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\:=\: \\ $$$$ \\ $$ Answered…
Question Number 149782 by iloveisrael last updated on 07/Aug/21 Answered by john_santu last updated on 07/Aug/21 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{6}{x}+\mathrm{2sin}\:\mathrm{6}{x}\mathrm{cos}\:\mathrm{4}{x}−\left(\mathrm{3sin}\:\mathrm{6}{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{6}{x}\right)}{\mathrm{3sin}\:{x}−\left(\mathrm{3sin}\:{x}−\mathrm{4sin}\:^{\mathrm{3}} {x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:\mathrm{6}{x}+\mathrm{2sin}\:\mathrm{6}{x}\:\mathrm{cos}\:\mathrm{4}{x}+\mathrm{4sin}\:^{\mathrm{3}} \mathrm{6}{x}\:}{\mathrm{4sin}\:^{\mathrm{3}} {x}}…
Question Number 18678 by Joel577 last updated on 27/Jul/17 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{4}^{{x}\:+\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:+\mathrm{1}} \:−\:\mathrm{3}^{{x}\:+\:\mathrm{1}} }{\mathrm{4}^{{x}\:−\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:−\:\mathrm{1}\:} +\:\mathrm{3}^{{x}\:+\:\mathrm{1}} \:} \\ $$ Answered by 433 last updated on…
Question Number 149700 by Lekhraj last updated on 06/Aug/21 Commented by iloveisrael last updated on 07/Aug/21 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{x}+\mathrm{x}\sqrt{\mathrm{1}+\mathrm{4x}^{−\mathrm{1}} }\right) \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}x}\left(\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}^{−\mathrm{1}} }\:\right) \\ $$$$=−\infty×\mathrm{2}\:=\:−\infty…
Question Number 18627 by Arnab Maiti last updated on 26/Jul/17 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\right)^{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}} \\ $$ Commented by Arnab Maiti last updated on 26/Jul/17 $$\mathrm{Please}\:\mathrm{help}. \\ $$…
Question Number 18611 by Arnab Maiti last updated on 25/Jul/17 $$\mathrm{Without}\:\mathrm{using}\:\mathrm{L}\:\mathrm{Hospital}'\mathrm{s} \\ $$$$\mathrm{rule}\:\mathrm{prove}\:\mathrm{that}\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}=\mathrm{1} \\ $$ Commented by Arnab Maiti last updated on 25/Jul/17 $$\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{NCERT}\:\mathrm{books}.…