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Question-162747




Question Number 162747 by tounghoungko last updated on 01/Jan/22
Answered by bobhans last updated on 01/Jan/22
  Happy New Year too   lim_(x→0)  ((sin 7x−7x+7x−7sin x)/(x^3 (((sin x)/x))^3 )) = lim_(x→0)  ((7(x−sin x))/x^3 )+7^3 .lim_(x→0)  ((sin 7x−7x)/((7x)^3 ))    = (7/6)−(7^3 /6) = ((7(1−49))/6) = 7×(−8)=−56
$$\:\:\mathrm{Happy}\:\mathrm{New}\:\mathrm{Year}\:\mathrm{too} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{7x}−\mathrm{7x}+\mathrm{7x}−\mathrm{7sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{3}} \left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\right)^{\mathrm{3}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{7}\left(\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{x}^{\mathrm{3}} }+\mathrm{7}^{\mathrm{3}} .\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{7x}−\mathrm{7x}}{\left(\mathrm{7x}\right)^{\mathrm{3}} } \\ $$$$\:\:=\:\frac{\mathrm{7}}{\mathrm{6}}−\frac{\mathrm{7}^{\mathrm{3}} }{\mathrm{6}}\:=\:\frac{\mathrm{7}\left(\mathrm{1}−\mathrm{49}\right)}{\mathrm{6}}\:=\:\mathrm{7}×\left(−\mathrm{8}\right)=−\mathrm{56} \\ $$

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