Question Number 203288 by hardmath last updated on 14/Jan/24
$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}\:−\:\mathrm{cos4x}}{\mathrm{x}^{\mathrm{2}} }\:=\:? \\ $$
Answered by MM42 last updated on 14/Jan/24
$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}{x}}{{x}^{\mathrm{2}} }\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\mathrm{8}\left(\frac{{sin}\mathrm{2}{x}}{\mathrm{2}{x}}\right)^{\mathrm{2}} =\mathrm{8}\:\checkmark \\ $$
Commented by hardmath last updated on 14/Jan/24
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$
Answered by Calculusboy last updated on 15/Jan/24
$$\boldsymbol{{Solution}}:\:\boldsymbol{{NB}}\:\boldsymbol{{that}}\:\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{1}−\boldsymbol{{cosax}}}{\boldsymbol{{x}}^{\mathrm{2}} }=\frac{\boldsymbol{{a}}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\boldsymbol{{then}};\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{1}−\boldsymbol{{cos}}\mathrm{4}\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{2}} }=\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{16}}{\mathrm{2}}=\mathrm{8} \\ $$$$\boldsymbol{{Mf}} \\ $$
Answered by dimentri last updated on 15/Jan/24
$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{4x}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{4x}\right)} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\mathrm{4x}}{\mathrm{x}}\right)^{\mathrm{2}} .\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\mathrm{4x}} \\ $$$$\:=\:\mathrm{16}×\frac{\mathrm{1}}{\mathrm{2}}=\:\mathrm{8} \\ $$