Question Number 91182 by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3tan}\:^{\mathrm{2}} {x}}{\mathrm{3}{x}^{\mathrm{6}} } \\ $$ Answered by john santu last updated on 28/Apr/20…
Question Number 91158 by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}}\:−\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{8}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} } \\ $$ Commented by jagoll last updated on 28/Apr/20 $${thank}\:{you}\:{sir} \\…
Question Number 91157 by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4}{x}−\mathrm{4tan}\:\mathrm{3}{x}}{\mathrm{3sin}\:\mathrm{4}{x}−\mathrm{4sin}\:\mathrm{3}{x}}\:=\:? \\ $$ Commented by john santu last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{4}{x}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{3}} }{\mathrm{3}}\right)−\mathrm{4}\left(\mathrm{3}{x}+\frac{\left(\mathrm{3}{x}\right)^{\mathrm{3}} }{\mathrm{3}}\right)}{\mathrm{3}\left(\mathrm{4}{x}−\frac{\left(\mathrm{4}{x}\right)^{\mathrm{3}}…
Question Number 91154 by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{12}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{12cos}\:{x}}{{x}^{\mathrm{4}} } \\ $$ Commented by jagoll last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}\left(\mathrm{2}−{x}^{\mathrm{2}} −\mathrm{2cos}\:{x}\right)}{{x}^{\mathrm{4}}…
Question Number 156683 by cortano last updated on 14/Oct/21 Commented by john_santu last updated on 14/Oct/21 $${it}\:{should}\:{be} \\ $$$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{1}+\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}−\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}{{x}−\mathrm{1}+\sqrt{{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:. \\…
Question Number 91133 by john santu last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left({x}−\mathrm{1}\right)+\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}−{x}}}{\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\: \\ $$ Commented by john santu last updated on 28/Apr/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left({x}−\mathrm{1}\right)+\left(\mathrm{1}−\frac{{x}}{\mathrm{3}}\right)}{\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{2}}\:\left(\mathrm{1}−\frac{{x}}{\mathrm{3}}\right)}\:=\:…
Question Number 25588 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{x}}−\boldsymbol{{tgx}}}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}\right)=? \\ $$$$\boldsymbol{{l}}'\boldsymbol{{hopital}}\:\boldsymbol{{role}}\:\boldsymbol{{not}}\:\boldsymbol{{allowed}}! \\ $$ Commented by moxhix last updated on 12/Dec/17 https://math.stackexchange.com/questions/508733/lim-x-to0-fracx-sin-xx-tan-x-without-using-lhopital . Answered by…
Question Number 156640 by semanou last updated on 13/Oct/21 Answered by puissant last updated on 13/Oct/21 $$\left.{c}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{{tanx}}=\:\mathrm{1} \\ $$$$\left.{d}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{{sinx}}=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 156611 by semanou last updated on 13/Oct/21 $${lim}\sqrt{{x}+\mathrm{5}} \\ $$ Commented by Rasheed.Sindhi last updated on 13/Oct/21 $${x}\rightarrow? \\ $$ Terms of Service…
Question Number 25495 by rita1608 last updated on 11/Dec/17 Commented by prakash jain last updated on 11/Dec/17 $$\mid{x}\mid=\begin{cases}{−{x}}&{{x}<\mathrm{0}}\\{{x}}&{{x}\geqslant\mathrm{0}}\end{cases} \\ $$$${f}\left({x}\right)=\begin{cases}{−{x}^{\mathrm{2}} }&{{x}<\mathrm{0}}\\{{x}^{\mathrm{2}} }&{{x}\geqslant\mathrm{0}}\end{cases} \\ $$$$\mathrm{clearly}\:\mathrm{it}\:\mathrm{differential}\:\mathrm{at}\:\mathrm{every}\:\mathrm{pt}\:\mathrm{other} \\…