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Question Number 86193 by john santu last updated on 27/Mar/20

lim_(x→0) ((4x^2 +((6x^2 )/(√(9x^4 +9sin^2 x))))/(3x−((4x^3 −x)/(2x+1)))) = ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4}{x}^{\mathrm{2}} +\frac{\mathrm{6}{x}^{\mathrm{2}} }{\sqrt{\mathrm{9}{x}^{\mathrm{4}} +\mathrm{9sin}\:^{\mathrm{2}} {x}}}}{\mathrm{3}{x}−\frac{\mathrm{4}{x}^{\mathrm{3}} −{x}}{\mathrm{2}{x}+\mathrm{1}}}\:=\:? \\ $$

Commented by jagoll last updated on 27/Mar/20

i like this question

$$\mathrm{i}\:\mathrm{like}\:\mathrm{this}\:\mathrm{question} \\ $$

Commented by MJS last updated on 27/Mar/20

the limit x→0^− is −(1/2) but the limit x→0^+ is +(1/2) ⇒ the limit doesn′t exist

$$\mathrm{the}\:\mathrm{limit}\:{x}\rightarrow\mathrm{0}^{−} \:\mathrm{is}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{but}\:\mathrm{the}\:\mathrm{limit}\:{x}\rightarrow\mathrm{0}^{+} \:\mathrm{is} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{the}\:\mathrm{limit}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$

Commented by MJS last updated on 28/Mar/20

f(x)=((4x^2 +((6x^2 )/(√(9x^4 +9sin^2 x))))/(3x−((4x^3 −x)/(2x+1))))<0 ∀x<0 just calculate f(−1), f(−(1/(10))), f(−(1/(100))), ...

$${f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} +\frac{\mathrm{6}{x}^{\mathrm{2}} }{\sqrt{\mathrm{9}{x}^{\mathrm{4}} +\mathrm{9sin}^{\mathrm{2}} \:{x}}}}{\mathrm{3}{x}−\frac{\mathrm{4}{x}^{\mathrm{3}} −{x}}{\mathrm{2}{x}+\mathrm{1}}}<\mathrm{0}\:\forall{x}<\mathrm{0} \\ $$$$\mathrm{just}\:\mathrm{calculate}\:{f}\left(−\mathrm{1}\right),\:{f}\left(−\frac{\mathrm{1}}{\mathrm{10}}\right),\:{f}\left(−\frac{\mathrm{1}}{\mathrm{100}}\right),\:... \\ $$

Commented by john santu last updated on 28/Mar/20

post your argument sir

$${post}\:{your}\:{argument}\:{sir} \\ $$

Commented by jagoll last updated on 28/Mar/20

why?

$$\mathrm{why}? \\ $$$$ \\ $$

Commented by john santu last updated on 28/Mar/20

Commented by john santu last updated on 28/Mar/20

this the graph?

$${this}\:{the}\:{graph}?\: \\ $$

Commented by MJS last updated on 28/Mar/20

the graph before or after cancelling and transforming the fraction? u(x)=4x^2 +((6x^2 )/(√(9x^4 +9sin^2 x))) u(x) is defined for x≠0 and u(x)>0∀x≠0 v(x)=3x−((4x^3 −x)/(2x+1)) v(x) is defined for x≠−(1/2) and { ((v(x)≥0; 0≤x≤2)),((v(x)<0; x<−(1/2)∨−(1/2)<x<0∨x>2)) :} ⇒ f(x)=((u(x))/(v(x))) is defined for x≠−(1/2)∧x≠0∧x≠2 and { ((f(x)≥0; 0<x<2)),((f(x)<0; x<−(1/2)∨−(1/2)<x<0∨x>2)) :}

$$\mathrm{the}\:\mathrm{graph}\:{before}\:\mathrm{or}\:{after}\:\mathrm{cancelling}\:\mathrm{and} \\ $$$$\mathrm{transforming}\:\mathrm{the}\:\mathrm{fraction}? \\ $$$${u}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} +\frac{\mathrm{6}{x}^{\mathrm{2}} }{\sqrt{\mathrm{9}{x}^{\mathrm{4}} +\mathrm{9sin}^{\mathrm{2}} \:{x}}} \\ $$$${u}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\neq\mathrm{0}\:\mathrm{and}\:{u}\left({x}\right)>\mathrm{0}\forall{x}\neq\mathrm{0} \\ $$$${v}\left({x}\right)=\mathrm{3}{x}−\frac{\mathrm{4}{x}^{\mathrm{3}} −{x}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$${v}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\neq−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\begin{cases}{{v}\left({x}\right)\geqslant\mathrm{0};\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}}\\{{v}\left({x}\right)<\mathrm{0};\:{x}<−\frac{\mathrm{1}}{\mathrm{2}}\vee−\frac{\mathrm{1}}{\mathrm{2}}<{x}<\mathrm{0}\vee{x}>\mathrm{2}}\end{cases} \\ $$$$\Rightarrow\:{f}\left({x}\right)=\frac{{u}\left({x}\right)}{{v}\left({x}\right)}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\neq−\frac{\mathrm{1}}{\mathrm{2}}\wedge{x}\neq\mathrm{0}\wedge{x}\neq\mathrm{2} \\ $$$$\mathrm{and}\:\begin{cases}{{f}\left({x}\right)\geqslant\mathrm{0};\:\mathrm{0}<{x}<\mathrm{2}}\\{{f}\left({x}\right)<\mathrm{0};\:{x}<−\frac{\mathrm{1}}{\mathrm{2}}\vee−\frac{\mathrm{1}}{\mathrm{2}}<{x}<\mathrm{0}\vee{x}>\mathrm{2}}\end{cases} \\ $$

Answered by john santu last updated on 27/Mar/20

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