Question Number 23765 by Anoop kumar last updated on 05/Nov/17 $$\underset{{x}\rightarrow\propto} {\mathrm{lim}}\:\frac{\mathrm{cot}^{−\mathrm{1}} \left({x}^{−{a}} \mathrm{ln}\:_{{a}} \:{x}\right)}{\mathrm{sec}^{−\mathrm{1}} \left({a}^{{x}} \:\mathrm{ln}\:_{{x}} {a}\right)}\:\: \\ $$$$\left({a}>\mathrm{1}\right) \\ $$$$ \\ $$ Terms…
Question Number 154776 by john_santu last updated on 21/Sep/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}−\frac{\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}}}{{e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} }\right)^{{x}} =? \\ $$ Commented by mathdanisur last updated on 21/Sep/21 $$\mathrm{say}\:\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{t}\:\Rightarrow\:\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\mathrm{e}^{\underset{\boldsymbol{\mathrm{t}}\rightarrow\mathrm{0}}…
Question Number 23688 by A1B1C1D1 last updated on 04/Nov/17 Answered by $@ty@m last updated on 04/Nov/17 $${Use}\:{method}\:{of}\:{rationalization} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{{x}^{\mathrm{2}} \left(\sqrt{{x}^{\mathrm{2}} +\mathrm{12}}+\sqrt{\mathrm{12}}\right)}{{x}^{\mathrm{2}} +\mathrm{12}−\mathrm{12}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 154719 by liberty last updated on 21/Sep/21 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2ln}\:\left[\Gamma\left(\mathrm{sin}\:{x}\right)\right]−\mathrm{ln}\:\pi}{\Gamma\left(\mathrm{sec}\:{x}\right)−\mathrm{1}}\:=? \\ $$ Answered by john_santu last updated on 21/Sep/21 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}\:\Psi^{\mathrm{0}} \left(\mathrm{sin}\:\left({x}\right)\right)}{\mathrm{2tan}\:\left(\mathrm{2}{x}\right)\mathrm{sec}\:\left(\mathrm{2}{x}\right)\Gamma\left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)\Psi^{\mathrm{0}} \left(\mathrm{sec}\:\left(\mathrm{2}{x}\right)\right)}= \\…
Question Number 23584 by A1B1C1D1 last updated on 02/Nov/17 Answered by FilupES last updated on 02/Nov/17 $$\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)}{{x}}=\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\right)\left(\underset{{x}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\left\{\mathrm{ln}\left(\mathrm{3}+{x}\right)−\mathrm{ln}\left({x}\right)\right\}\right) \\ $$$$\:…
Question Number 154652 by EDWIN88 last updated on 20/Sep/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\mathrm{tan}\:\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{4}}\right)\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{3}{x}\right)}{{x}^{\mathrm{3}} }=? \\ $$ Answered by bramlexs22 last updated on 20/Sep/21 Commented by mnjuly1970 last…
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Question Number 154593 by Niiicooooo last updated on 19/Sep/21 Commented by Niiicooooo last updated on 19/Sep/21 Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21 $${S}=\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 154586 by Dadoubleny last updated on 19/Sep/21 Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\left({n}+\mathrm{1}\right)^{\alpha} −{n}^{\alpha} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({n}^{\alpha} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{\alpha} −{n}^{\alpha}…
Question Number 88985 by jagoll last updated on 14/Apr/20 $$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{ln}\:\mid{x}−\mathrm{3}\mid\: \\ $$$${exist}\:{or}\:{no}? \\ $$ Commented by mr W last updated on 14/Apr/20 $${no}. \\…