Question Number 66927 by Cmr 237 last updated on 21/Aug/19 $${calcul}\:{la}\:{limite}\:{suivante}: \\ $$$${lim}\:\:\:\:\:\:\:\left(\frac{\mathrm{1}^{{x}} +\mathrm{2}^{{x}} +\mathrm{3}^{{x}} +……+{n}^{{x}} }{{n}}\right)^{\frac{\mathrm{1}}{{x}}} =\mathrm{A} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$\mathrm{trouve}\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{de}\:\boldsymbol{\mathrm{A}} \\ $$$$\mathrm{trouve}\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{de}\:\boldsymbol{\mathrm{P}}\:\mathrm{definir}\:\mathrm{par}: \\…
Question Number 1368 by prakash jain last updated on 25/Jul/15 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{ln}\left({ax}+{b}\right)}{\mathrm{ln}\:\left({bx}+{a}\right)}\:=? \\ $$ Commented by 123456 last updated on 25/Jul/15 $${f}\left({x}\right)=\frac{\mathrm{ln}\:\left({ax}+{b}\right)}{\mathrm{ln}\:\left({bx}+{a}\right)} \\ $$$$\frac{\frac{{d}}{{dx}}\left[\mathrm{ln}\:\left({ax}+{b}\right)\right]}{\frac{{d}}{{dx}}\left[\mathrm{ln}\:\left({bx}+{a}\right)\right]}=\frac{\frac{{a}}{{ax}+{b}}}{\frac{{b}}{{bx}+{a}}}=\frac{{a}\left({bx}+{a}\right)}{{b}\left({ax}+{b}\right)}\rightarrow\frac{{ab}}{{ba}}=\mathrm{1} \\…
Question Number 66862 by Aman Arya last updated on 20/Aug/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}3}{x}^{\mathrm{5}} \\ $$ Commented by mathmax by abdo last updated on 20/Aug/19 $$=\mathrm{3}×\mathrm{0}=\mathrm{0} \\…
Question Number 132367 by Raxreedoroid last updated on 13/Feb/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{x}^{{n}} }{\Gamma\left({n}+\mathrm{1}\right)} \\ $$ Answered by TheSupreme last updated on 13/Feb/21 $${if}\:{n}\in\mathbb{N}\:\rightarrow\:\Gamma\left({n}+\mathrm{1}\right)={n}! \\ $$$$\mathrm{lim}\:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}=\mathrm{0}\:\forall{x}\in\mathbb{R}_{\mathrm{0}}…
Question Number 132356 by Raxreedoroid last updated on 13/Feb/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}^{{n}} }{{n}!}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{ln}\:\mathrm{2}\right)^{{n}} }\right)=?\:,{x}\in\mathbb{R} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 132328 by liberty last updated on 13/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{p}−\mathrm{x}−\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} −…−\mathrm{x}^{\mathrm{p}} }{\mathrm{x}−\mathrm{1}}\:=? \\ $$ Answered by EDWIN88 last updated on 13/Feb/21 $$\:\mathrm{The}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{form}\:\frac{\mathrm{0}}{\mathrm{0}} \\…
Question Number 132321 by Raxreedoroid last updated on 13/Feb/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{cos}\:\left({x}\mathrm{ln}\:{k}\right)}{\left(\mathrm{ln}\:{k}\right)^{{n}} \sqrt{{k}}}=? \\ $$$$\mathrm{where}\:{k},{n}\:\in\mathbb{N}\:,\:{x}\in\mathbb{R} \\ $$ Answered by Dwaipayan Shikari last updated on 13/Feb/21 $$\underset{{n}\rightarrow\infty}…
Question Number 1200 by sumitkumar4799@gmail.com last updated on 14/Jul/15 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\mathrm{cos}{x}.\mathrm{sin}{x}\right] \\ $$ Answered by prakash jain last updated on 14/Jul/15 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\right]=\mathrm{0} \\ $$…
Question Number 132240 by bemath last updated on 12/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{2tan}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{tan}\:\frac{\pi}{\mathrm{4}}}{\mathrm{sin}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{2sin}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}}\:? \\ $$$$ \\ $$ Answered by EDWIN88 last updated on 13/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sec}\:^{\mathrm{2}} \left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{2sec}\:^{\mathrm{2}}…
Question Number 132226 by kaivan.ahmadi last updated on 12/Feb/21 $${li}\underset{{x}\rightarrow\mathrm{0}^{+} } {{m}}\:\:{xln}\left({sinx}\right) \\ $$ Answered by Olaf last updated on 12/Feb/21 $$\mathrm{sin}{x}\:\underset{\mathrm{0}} {\sim}\:{x} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}}…