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Category: Limits

lim-x-0-1-mx-1-nx-mn-x-

Question Number 80455 by jagoll last updated on 03/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+{mx}}{\mathrm{1}−{nx}}\right)^{\frac{{mn}}{{x}}} \\ $$ Commented by mr W last updated on 03/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+{mx}}{\mathrm{1}−{nx}}\right)^{\frac{{mn}}{{x}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}}…

lim-x-0-1-x-1-x-e-x-

Question Number 80417 by jagoll last updated on 03/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}}{{x}}\:=\:? \\ $$ Commented by MJS last updated on 03/Feb/20 $$\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{approximate}\:\mathrm{and}\:\mathrm{got}\:\mathrm{something} \\ $$$$\mathrm{very}\:\mathrm{close}\:\mathrm{to}\:−\frac{\mathrm{e}}{\mathrm{2}}\:\mathrm{but}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{prove}\:\mathrm{the} \\…

n-1-1-n-n-

Question Number 145951 by ArielVyny last updated on 09/Jul/21 $$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}=?? \\ $$ Answered by Olaf_Thorendsen last updated on 09/Jul/21 $$\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\:\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +…\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty}…

lim-x-0-ln-tan-x-1-sin-x-xsin-x-

Question Number 80343 by jagoll last updated on 02/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ln}\left(\mathrm{tan}\:{x}+\mathrm{1}\right)−\mathrm{sin}\:{x}}{{x}\mathrm{sin}\:{x}} \\ $$ Commented by abdomathmax last updated on 02/Feb/20 $${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}=\mathrm{1}−{u}\:+{o}\left({u}^{\mathrm{2}} \right)\left({u}\sim\mathrm{0}\right) \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{u}\right)={u}−\frac{{u}^{\mathrm{2}}…

lim-x-x-tan-1-x-1-x-4-pi-4-

Question Number 145806 by bramlexs22 last updated on 08/Jul/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}+\mathrm{4}}\right)−\frac{\pi}{\mathrm{4}}\right)\:=?\: \\ $$ Answered by gsk2684 last updated on 08/Jul/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{{x}+\mathrm{1}}{{x}+\mathrm{4}}−\mathrm{1}}{\mathrm{1}+\frac{{x}+\mathrm{1}}{{x}+\mathrm{4}}}\right)\right) \\…