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
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}\left(\mathrm{2}−{x}^{\mathrm{2}} −\mathrm{2cos}\:{x}\right)}{{x}^{\mathrm{4}} }\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}\left(−\mathrm{2}{x}+\mathrm{2sin}\:{x}\right)}{\mathrm{4}{x}^{\mathrm{3}} }\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}\left(−\mathrm{2}+\mathrm{2cos}\:{x}\right)}{\mathrm{12}{x}^{\mathrm{2}} }\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:{x}}{\mathrm{4}{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by john santu last updated on 28/Apr/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{12}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{12}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{24}}\right)}{{x}^{\mathrm{4}} }\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{12}−\mathrm{6}{x}^{\mathrm{2}} −\mathrm{12}+\mathrm{6}{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{4}} }\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{4}} }\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$