lim-x-0-4x-2-6x-2-9x-4-9sin-2-x-3x-4x-3-x-2x-1-

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}), \dots

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 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