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Question Number 127002 by bramlexs22 last updated on 26/Dec/20

 lim_(x→0)  ((2+cos (3x)−3csch (x))/(ln (1+x^2 ))) =?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}+\mathrm{cos}\:\left(\mathrm{3}{x}\right)−\mathrm{3csch}\:\left({x}\right)}{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=? \\ $$

Commented by Evimene last updated on 26/Dec/20

      add me on facebook  on cadet paise maldonna just not used to  it here my data will soon finish

$$ \\ $$$$\:\: \\ $$$$\mathrm{add}\:\mathrm{me}\:\mathrm{on}\:\mathrm{facebook} \\ $$$$\mathrm{on}\:\mathrm{cadet}\:\mathrm{paise}\:\mathrm{maldonna}\:\mathrm{just}\:\mathrm{not}\:\mathrm{used}\:\mathrm{to} \\ $$$$\mathrm{it}\:\mathrm{here}\:\mathrm{my}\:\mathrm{data}\:\mathrm{will}\:\mathrm{soon}\:\mathrm{finish} \\ $$

Answered by liberty last updated on 26/Dec/20

 lim_(x→0)  ((2+1−((9x^2 )/2)−3(1+(x^2 /2)))/x^2 ) =   lim_(x→0)  ((3−((9x^2 )/2)−3−((3x^2 )/2))/x^2 ) = lim_(x→0)  ((−6x^2 )/x^2 ) = −6.

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}+\mathrm{1}−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{3}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{3}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}}{{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{6}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:=\:−\mathrm{6}. \\ $$

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