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Question-122776




Question Number 122776 by shaker last updated on 19/Nov/20
Answered by mathmax by abdo last updated on 19/Nov/20
without using anything....!
$$\mathrm{without}\:\mathrm{using}\:\mathrm{anything}….! \\ $$
Answered by kaivan.ahmadi last updated on 19/Nov/20
lim_(x→0) (((2x^4 )/(3x^6 )))+(((3x^2 )/(3x^6 )))−(((3tg^2 x)/(3x^6 )))=  lim_(x→0)  (2/(3x^2 ))+(1/x^4 )−(1/x^4 )=lim_(x→0) (2/(3x^2 ))=+∞
$${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{3}{x}^{\mathrm{6}} }\right)+\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}{x}^{\mathrm{6}} }\right)−\left(\frac{\mathrm{3}{tg}^{\mathrm{2}} {x}}{\mathrm{3}{x}^{\mathrm{6}} }\right)= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }={lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} }=+\infty \\ $$

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