Question Number 149959 by Mathfinity last updated on 08/Aug/21
Commented by john_santu last updated on 08/Aug/21
Commented by john_santu last updated on 08/Aug/21
Commented by john_santu last updated on 08/Aug/21
$$ \\ $$why do you take mr iloveisrael's question?
Answered by gsk2684 last updated on 08/Aug/21
$$\left.{i}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}{x}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)+\left(\mathrm{6}{x}−\frac{\left(\mathrm{6}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)+\left(\mathrm{10}{x}−\frac{\left(\mathrm{10}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)−\left(\mathrm{18}{x}−\frac{\left(\mathrm{18}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)}{\mathrm{3}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+..\right)−\left(\mathrm{3}{x}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}+..\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{−\mathrm{8}−\mathrm{216}−\mathrm{1000}+\mathrm{5832}}{\mathrm{6}}{x}^{\mathrm{3}} +..}{\frac{−\mathrm{3}+\mathrm{27}}{\mathrm{6}}{x}^{\mathrm{3}} +..} \\ $$$$=\frac{\frac{−\mathrm{8}−\mathrm{216}−\mathrm{1000}+\mathrm{5832}}{\mathrm{6}}}{\frac{−\mathrm{3}+\mathrm{27}}{\mathrm{6}}} \\ $$$$=\frac{\mathrm{4608}}{\mathrm{24}}=\mathrm{192} \\ $$$${or} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{6}{x}+\mathrm{2}{x}}{\mathrm{2}}\:\mathrm{cos}\frac{\mathrm{6}{x}−\mathrm{2}{x}}{\mathrm{2}}\:+\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{10}{x}+\mathrm{18}{x}}{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{10}{x}−\mathrm{18}{x}}{\mathrm{2}}}{\mathrm{4}\:\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{4}{x}.\:\mathrm{cos}\:\mathrm{2}{x}\:−\mathrm{2}\:\mathrm{cos}\:\mathrm{14}{x}.\:\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{4}\:\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{4}{x}\left(\:\mathrm{cos}\:\mathrm{2}{x}\:−\:\mathrm{cos}\:\mathrm{14}{x}\right)}{\mathrm{4}\:\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{4}{x}.\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{14}{x}+\mathrm{2}{x}}{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{14}{x}−\mathrm{2}{x}}{\mathrm{2}}}{\mathrm{4}\:\mathrm{sin}\:^{\mathrm{3}} {x}}\:\: \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\:\mathrm{sin}\:\mathrm{4}{x}.\:\mathrm{sin}\:\mathrm{8}{x}.\:\mathrm{sin}\:\mathrm{6}{x}}{\:\mathrm{sin}\:^{\mathrm{3}} {x}}\:\: \\ $$$$=\mathrm{4}×\mathrm{8}×\mathrm{6}=\mathrm{192} \\ $$$$\left.{ii}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}+\mathrm{sin}\:{x}\right)}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\underset{\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)^{\mathrm{2}} }\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\frac{\pi}{\mathrm{4}}{x}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$=\left(\mathrm{1}\right)^{\mathrm{2}} \underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\right)^{\mathrm{2}} =\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$ \\ $$
Commented by john_santu last updated on 08/Aug/21
$${false}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{sin}\:{x}}{\mathrm{2}{x}}\right)^{\mathrm{2}} \neq\:\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$