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lim-x-3x-tan-2-x-2x-sin-3-x-cos-1-x-cos-2-x-




Question Number 174492 by naka3546 last updated on 02/Aug/22
lim_(x→∞)   ((3x tan (2/x) − 2x sin (3/x))/(cos (1/x) − cos (2/x))) =  ?
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}{x}\:\mathrm{tan}\:\frac{\mathrm{2}}{{x}}\:−\:\mathrm{2}{x}\:\mathrm{sin}\:\frac{\mathrm{3}}{{x}}}{\mathrm{cos}\:\frac{\mathrm{1}}{{x}}\:−\:\mathrm{cos}\:\frac{\mathrm{2}}{{x}}}\:=\:\:? \\ $$
Commented by kaivan.ahmadi last updated on 02/Aug/22
we use Taylor series  ∼lim_(x→∞)  ((3x((2/x)−(8/(3x^3 )))−2x((3/x)−(9/(2x^3 ))))/((1−(1/(2x^2 ))+(1/(24x^4 )))−(1−(1/(2x^2 ))+(2/(3x^4 )))))    =lim_(x→∞)  ((−(8/x^2 )+(9/x^2 ))/((1/(24x^4 ))−(2/(3x^4 ))))=    lim_(x→∞)  ((1/x^2 )/((−15)/(24x^4 )))=lim_(x→∞)  ((24x^2 )/(−15))=−∞
$${we}\:{use}\:{Taylor}\:{series} \\ $$$$\sim{li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\mathrm{3}{x}\left(\frac{\mathrm{2}}{{x}}−\frac{\mathrm{8}}{\mathrm{3}{x}^{\mathrm{3}} }\right)−\mathrm{2}{x}\left(\frac{\mathrm{3}}{{x}}−\frac{\mathrm{9}}{\mathrm{2}{x}^{\mathrm{3}} }\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{24}{x}^{\mathrm{4}} }\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{4}} }\right)} \\ $$$$ \\ $$$$={li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{−\frac{\mathrm{8}}{{x}^{\mathrm{2}} }+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }}{\frac{\mathrm{1}}{\mathrm{24}{x}^{\mathrm{4}} }−\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{4}} }}= \\ $$$$ \\ $$$${li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{\frac{−\mathrm{15}}{\mathrm{24}{x}^{\mathrm{4}} }}={li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\mathrm{24}{x}^{\mathrm{2}} }{−\mathrm{15}}=−\infty \\ $$$$ \\ $$$$ \\ $$

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