Question Number 182560 by universe last updated on 11/Dec/22 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\mathrm{cot}^{\mathrm{2}} {x}\:=\:\:? \\ $$ Answered by cortano1 last updated on 11/Dec/22 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\:.\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 116970 by bemath last updated on 08/Oct/20 Answered by MJS_new last updated on 08/Oct/20 $$\mathrm{the}\:\mathrm{limit}\:{x}\rightarrow\mathrm{1}\:\mathrm{is}\:\mathrm{just}\:\mathrm{the}\:\mathrm{value}?! \\ $$$$=\sqrt{\frac{\mathrm{7}}{\mathrm{3}}} \\ $$ Commented by bemath last…
Question Number 116930 by bobhans last updated on 08/Oct/20 $$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{5}^{\sqrt{\mathrm{x}}} \:\mathrm{find}\:\mathrm{f}\:'\left(\mathrm{x}\right)\:\mathrm{by}\:\mathrm{using}\:\mathrm{limit} \\ $$$$\left(\mathrm{first}\:\mathrm{principal}\:\mathrm{derivative}\right) \\ $$ Commented by bobhans last updated on 08/Oct/20 Answered by 1549442205PVT…
Question Number 182453 by universe last updated on 09/Dec/22 Answered by qaz last updated on 10/Dec/22 $$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cot}\:{x}−\mathrm{cot}\:\mathrm{2}{x} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\mathrm{tan}\:\frac{\Psi}{\mathrm{2}^{{k}} }=\mathrm{tan}\:\Psi+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}^{{k}}…
Question Number 182441 by mathlove last updated on 09/Dec/22 Answered by JDamian last updated on 09/Dec/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{xk}} \right)−\mathrm{ln}\:{n}}{{x}}=\mathrm{9} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 116872 by bemath last updated on 07/Oct/20 Answered by bobhans last updated on 07/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{x}}\right)}{\mathrm{3}\:\mathrm{tan}\:\mathrm{2x}} \\ $$$$\mathrm{letting}\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{z}\:\mathrm{with}\:\mathrm{z}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{z}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{5z}\right)}{\mathrm{3}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{z}}\right)}\:=\:\underset{\mathrm{z}\rightarrow\mathrm{0}}…
Question Number 116859 by Study last updated on 07/Oct/20 $${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}^{{x}} }=? \\ $$ Commented by JDamian last updated on 07/Oct/20 As exponential function increases more rapidly than parabola, you even can get this limit intuitively Answered by…
Question Number 182346 by cortano1 last updated on 08/Dec/22 Commented by SANOGO last updated on 08/Dec/22 $${thank}\:{you} \\ $$ Answered by CrispyXYZ last updated on…
Question Number 116797 by ravisoni505 last updated on 06/Oct/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116756 by bemath last updated on 06/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{x}−\mathrm{2}}}{\mathrm{x}−\mathrm{2}}\:=? \\ $$ Answered by bobhans last updated on 06/Oct/20 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{x}−\mathrm{2}}}{\mathrm{x}−\mathrm{2}}=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} }}\:=\:\infty \\…