Question Number 116872 by bemath last updated on 07/Oct/20 | ||
Answered by bobhans last updated on 07/Oct/20 | ||
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{x}}\right)}{\mathrm{3}\:\mathrm{tan}\:\mathrm{2x}} \\ $$$$\mathrm{letting}\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{z}\:\mathrm{with}\:\mathrm{z}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{z}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{5z}\right)}{\mathrm{3}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{z}}\right)}\:=\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{z}\:\mathrm{cot}\:\:\left(\frac{\mathrm{1}}{\mathrm{z}}\right)\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{5z}\right) \\ $$$$\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{z}\:\mathrm{cot}\:\left(\frac{\mathrm{1}}{\mathrm{z}}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$ | ||