Question Number 99355 by bemath last updated on 20/Jun/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:−\mathrm{cot}\:^{\mathrm{2}} \mathrm{x}\:\right)\:=? \\ $$ Commented by john santu last updated on 20/Jun/20 $$\mathrm{maybe}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…
Question Number 99350 by bemath last updated on 20/Jun/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3}−\mathrm{x}}{\:\sqrt{\mathrm{9x}^{\mathrm{2}} −\mathrm{8}}\:}\:? \\ $$ Commented by bobhans last updated on 20/Jun/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{−\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}}\right)}{−\mathrm{x}\sqrt{\mathrm{9}−\frac{\mathrm{8}}{\mathrm{x}^{\mathrm{2}} }}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\…
Question Number 99348 by bobhans last updated on 20/Jun/20 Commented by bemath last updated on 20/Jun/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3x}+\sqrt{\mathrm{x}^{\mathrm{2}} }}{\mathrm{7x}−\mathrm{5}\sqrt{\mathrm{x}^{\mathrm{2}} }}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3x}−\mathrm{x}}{\mathrm{7x}+\mathrm{5x}} \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{2x}}{\mathrm{12x}}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\…
Question Number 99344 by bemath last updated on 20/Jun/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}^{\mathrm{x}} \:+\:\mathrm{3}^{\mathrm{x}} \:\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:? \\ $$ Answered by abdomsup last updated on 20/Jun/20 $${let}\:{f}\left({x}\right)=\left(\mathrm{2}^{{x}} \:+\mathrm{3}^{{x}}…
Question Number 99341 by bemath last updated on 20/Jun/20 $$\mathrm{if}\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{x}−\sqrt{\mathrm{ax}+\mathrm{b}}}{\mathrm{2x}−\mathrm{6}}\:=\:−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{then}\:\mathrm{a}−\mathrm{2b}\:=\: \\ $$ Commented by bobhans last updated on 20/Jun/20 $$\left(\mathrm{1}\right)\:\mathrm{3}−\sqrt{\mathrm{3a}+\mathrm{b}}\:=\:\mathrm{0}\: \\ $$$$\mathrm{9}\:=\:\mathrm{3a}+\mathrm{b}\:\Rightarrow\mathrm{b}\:=\:\mathrm{9}−\mathrm{3a}…
Question Number 99340 by bobhans last updated on 20/Jun/20 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{2x}−\mathrm{a}\:,\:\mathrm{x}<−\mathrm{3}}\\{\mathrm{ax}+\mathrm{b}\:,\:−\mathrm{3}\leqslant\mathrm{x}\leqslant\mathrm{3}}\\{\mathrm{b}−\mathrm{5x}\:,\:\mathrm{x}>\mathrm{3}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}f}\left(\mathrm{x}\right)\:\mathrm{and}\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{exist}\: \\ $$ Commented by PRITHWISH SEN 2 last updated…
Question Number 99262 by Ar Brandon last updated on 19/Jun/20 $$\mathrm{Consider}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{xe}^{\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{Find}\:\mathrm{3}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\:\mathrm{b},\:\mathrm{and}\:\mathrm{c}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{ax}+\mathrm{b}+\frac{\mathrm{c}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}}\xi\left(\mathrm{x}\right) \\ $$$$\mathrm{where}\:\underset{\mathrm{x}\rightarrow\pm\infty} {\mathrm{lim}}\xi\left(\mathrm{x}\right)=\mathrm{0} \\ $$ Terms of Service Privacy…
Question Number 99244 by Ar Brandon last updated on 19/Jun/20 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}−\mathrm{4}}−\mathrm{2}}{\mathrm{2x}}=? \\ $$ Answered by mahdi last updated on 19/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}+\mathrm{4}}−\mathrm{2}}{\mathrm{2x}}×\frac{\sqrt{\mathrm{x}+\mathrm{4}}+\mathrm{2}}{\:\sqrt{\mathrm{x}−\mathrm{4}}+\mathrm{2}}= \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 99193 by bemath last updated on 19/Jun/20 Commented by I want to learn more last updated on 19/Jun/20 $$\mathrm{Sir},\:\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{type}\:\mathrm{like}\:\mathrm{this} \\ $$ Commented by…
Question Number 164705 by Kayela last updated on 20/Jan/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}^{{x}} +\mathrm{2}^{{x}} +…+{n}^{{x}} }{{n}}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ Answered by Berlindo last updated on 20/Jan/22 $$=\left[\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}^{\mathrm{x}}…