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Question Number 215343 by RoseAli last updated on 03/Jan/25

lim_(x→0) ((sin 3xcos5x )/x^2 )

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{3}{x}\mathrm{cos5}{x}\:}{{x}^{\mathrm{2}} } \\ $$

Commented by Frix last updated on 04/Jan/25

Does not exist.

$$\mathrm{Does}\:\mathrm{not}\:\mathrm{exist}. \\ $$

Answered by MrGaster last updated on 04/Jan/25

lim_(x→0) ((sin 3xcos5x )/x^2 ) lim_(x→0) (((sin 3x)/(3x))∙((3 cos5x)/(2x))) =lim_(x→0) (((sin 3x)/(3x)))∙lim_(x→0) (((3 cos5x)/(2x))) =1∙lim_(x→0) (((3 cos5x)/(2x))) =lim_(x→0) (((3 cos5x)/(2x))) =lim_(x→0) (((3 cos5x)/2)∙(1/x)) =(3/2)lim_(x→0) (((cos5x)/x)) =(3/2)∙lim_(x→0) (((cos5x−cos5∙0)/(x−0))) =(3/2)∙(−5 sin5∙0) =(3/2)∙(−5∙0) =0 or −15

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{3}{x}\mathrm{cos5}{x}\:}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{3}{x}}\centerdot\frac{\mathrm{3}\:\mathrm{cos5}{x}}{\mathrm{2}{x}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{3}{x}}\right)\centerdot\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}\:\mathrm{cos5}{x}}{\mathrm{2}{x}}\right) \\ $$$$=\mathrm{1}\centerdot\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}\:\mathrm{cos5}{x}}{\mathrm{2}{x}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}\:\mathrm{cos5}{x}}{\mathrm{2}{x}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}\:\mathrm{cos5}{x}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos5}{x}}{{x}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\centerdot\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos5}{x}−\mathrm{cos5}\centerdot\mathrm{0}}{{x}−\mathrm{0}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\centerdot\left(−\mathrm{5}\:\mathrm{sin5}\centerdot\mathrm{0}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\centerdot\left(−\mathrm{5}\centerdot\mathrm{0}\right) \\ $$$$=\mathrm{0}\:\mathrm{or}\:−\mathrm{15} \\ $$

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