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 $$\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{NCERT}\:\mathrm{books}.…
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Question Number 84130 by mahdi last updated on 09/Mar/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}x}\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}}−\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}+\mathrm{x}\right) \\ $$ Commented by MJS last updated on 10/Mar/20 $$\mathrm{let}\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}^{+}…
Question Number 18593 by Joel577 last updated on 25/Jul/17 Answered by 433 last updated on 25/Jul/17 $$ \\ $$$$ \\ $$$$ \\ $$$$\frac{\mathrm{2016}\left(\frac{\mathrm{1}^{\mathrm{2015}} }{{n}^{\mathrm{2014}} }+\frac{\mathrm{2}^{\mathrm{2015}}…
Question Number 149660 by fotosy2k last updated on 06/Aug/21 Answered by JDamian last updated on 06/Aug/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{a}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }+\frac{{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }\right)= \\…
Question Number 18587 by Joel577 last updated on 25/Jul/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{{x}\:\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\:+\:\mathrm{1}} \\ $$ Commented by Joel577 last updated on 25/Jul/17 $$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{{x}\:\mathrm{sin}\:{x}\:+\:\mathrm{2sin}^{\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{2}}{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{\mathrm{sin}\:{x}\:\left({x}\:+\:\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)} \\…