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


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

$\lim_{x \to 0} \frac{4 x^{2} + \frac{6 x^{2}}{\sqrt{9 x^{4} + 9 \sin^{2} x}}}{3 x - \frac{4 x^{3} - x}{2 x + 1}} = ?$
Commented by jagoll last updated on 27/Mar/20

$i like this question$
Commented by MJS last updated on 27/Mar/20

$the limit x \to 0^{-} is - \frac{1}{2} but the limit x \to 0^{+} is$ $+ \frac{1}{2} \Rightarrow the limit {doesn}^{'} t exist$
Commented by MJS last updated on 28/Mar/20

$f (x) = \frac{4 x^{2} + \frac{6 x^{2}}{\sqrt{9 x^{4} + {9 \sin}^{2} x}}}{3 x - \frac{4 x^{3} - x}{2 x + 1}} < 0 \forall x < 0$ $just calculate f (- 1), f (- \frac{1}{10}), f (- \frac{1}{100}), . . .$
Commented by john santu last updated on 28/Mar/20

$p o s t y o u r a r g u m e n t s i r$
Commented by jagoll last updated on 28/Mar/20

$why ?$
Commented by john santu last updated on 28/Mar/20

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

$t h i s t h e g r a p h ?$
Commented by MJS last updated on 28/Mar/20
$the graph before or after cancelling and transforming the fraction? u(x)=4x^2 +((6x^2 )/(√(9x^4 +9sin^2 x))) u(x) is defined for x≠0 and u(x)>0∀x≠0 v(x)=3x−((4x^3 −x)/(2x+1)) v(x) is defined for x≠−(1/2) and { ((v(x)≥0; 0≤x≤2)),((v(x)<0; x<−(1/2)∨−(1/2)<x<0∨x>2)) :} ⇒ f(x)=((u(x))/(v(x))) is defined for x≠−(1/2)∧x≠0∧x≠2 and { ((f(x)≥0; 0<x<2)),((f(x)<0; x<−(1/2)∨−(1/2)<x<0∨x>2)) :}$
$the graph b e f o r e or a f t e r cancelling and$ $transforming the fraction ?$ $u (x) = 4 x^{2} + \frac{6 x^{2}}{\sqrt{9 x^{4} + {9 \sin}^{2} x}}$ $u (x) is defined for x \neq 0 and u (x) > 0 \forall x \neq 0$ $v (x) = 3 x - \frac{4 x^{3} - x}{2 x + 1}$ $v (x) is defined for x \neq - \frac{1}{2} and {\begin{cases} v (x) ⩾ 0; 0 ⩽ x ⩽ 2 \\ v (x) < 0; x < - \frac{1}{2} \lor - \frac{1}{2} < x < 0 \lor x > 2 \end{cases}$ $\Rightarrow f (x) = \frac{u (x)}{v (x)} is defined for x \neq - \frac{1}{2} \land x \neq 0 \land x \neq 2$ $and {\begin{cases} f (x) ⩾ 0; 0 < x < 2 \\ f (x) < 0; x < - \frac{1}{2} \lor - \frac{1}{2} < x < 0 \lor x > 2 \end{cases}$
	Answered by john santu last updated on 27/Mar/20

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