lim-x-1-ax-b-1-3x-2-c-2a-2b-3c-a-b-c-0-pease-solution-

Question Number 184773 by Ml last updated on 11/Jan/23

\lim_{x \to 1} \frac{ax + b}{\sqrt{1 + 3 x} - 2} = c

2 a - 2 b + 3 c = ?

(a, b, c) \neq 0

pease solution ? ? ? ?

Commented by SEKRET last updated on 11/Jan/23

2 a - 2 b - 3 c = 0

Answered by SEKRET last updated on 11/Jan/23

\lim_{x \to 1} \frac{ax + b}{\sqrt{1 + 3 x} - 2} = c

a = const b = const c = const

a = - b

\lim_{x \to 1} \frac{ax - a}{\sqrt{1 + 3 x} - 2} = c

\lim_{x \to 0} \frac{a (x - 1) \cdot (\sqrt{1 + 3 x} + 2)}{3 (x - 1)} = c

4 a = 3 c = - 4 b

2 a - 2 b + 3 c = 2 a + 2 a + 3 c = 4 a + 3 c = 6 c

2 a - 2 b + 3 c = 6 c = 8 a = - 8 b

Answered by a.lgnaoui last updated on 12/Jan/23

\frac{a x + b}{\sqrt{1 + 3 x} - 2} = \frac{(a x + b) (\sqrt{1 + 3 x} + 2)}{x - 1} = 3 c

\sqrt{1 + 3 x} = \frac{3 c (x - 1)}{a x + b} - 2

1 + 3 x = 4 + \frac{9 c^{2} {(x - 1)}^{2}}{{(a x + b)}^{2}} - \frac{12 c (x - 1) (a x + b)}{{(a x + b)}^{2}}

(1 + 3 x) {(a x + b)}^{2} = 4 {(a x + b)}^{2} - 12 c (x - 1) (a x + b) + 9 c^{2} {(x - 1)}^{2}

3 x {(a x + b)}^{2} = 3 {(a x + b)}^{2} - 12 c (a x + b) (x - 1) + 9 c^{2} {(x - 1)}^{2}

3 (x - 1) {(a x + b)}^{2} + 12 (x - 1) (a x + b) - 9 c^{2} {(x - 1)}^{2} = 0

{(a x + b)}^{2} + 4 c (a x + b) - 3 c^{2} (x - 1)

{(a x + b + 2 c)}^{2} - c^{2} (1 + 3 x) = 0

c \neq 0

4 = {(\frac{a + b + 2 c}{c})}^{2} \Rightarrow \frac{a + b + 2 c}{c} = 2

{\begin{cases} c = R - {0} \\ a = - b e t (a, b) \in R^{2} - {0, 0} \end{cases}