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

$$\underset{x\to 1}{\lim }\frac{\mathrm{ax}+\mathrm{b}}{\sqrt{1+3\mathrm{x}}-2}=\mathrm{c}$$$$2\mathrm{a}-2\mathrm{b}+3\mathrm{c}=?$$$$(\mathrm{a},\mathrm{b},\mathrm{c})\ne 0$$$$\mathrm{pease}\mathrm{solution}????$$

Commented by SEKRET last updated on 11/Jan/23

$$2\mathbf{a}-2\mathbf{b}-3\mathbf{c}=0$$

Answered by SEKRET last updated on 11/Jan/23

$${\mathbf{lim}}_{\mathbf{x}\to 1}\frac{\mathbf{ax}+\mathbf{b}}{\sqrt{1+3\mathbf{x}}-2}=\mathbf{c}$$$$\mathbf{a}=\mathbf{const}\mathbf{b}=\mathbf{const}\mathbf{c}=\mathbf{const}$$$$\mathbf{a}=-\mathbf{b}$$$${\mathbf{lim}}_{\mathbf{x}\to 1}\frac{\mathbf{ax}-\mathbf{a}}{\sqrt{1+3\mathbf{x}}-2}=\mathbf{c}$$$${\mathbf{lim}}_{\mathbf{x}\to 0}\frac{\mathbf{a}(\mathbf{x}-1)\cdot (\sqrt{1+3\mathbf{x}}+2)}{3(\mathbf{x}-1)}=\mathbf{c}$$$$4\mathbf{a}=3\mathbf{c}=-4\mathbf{b}$$$$2\mathbf{a}-2\mathbf{b}+3\mathbf{c}=2\mathbf{a}+2\mathbf{a}+3\mathbf{c}=4\mathbf{a}+3\mathbf{c}=6\mathbf{c}$$$$2\mathbf{a}-2\mathbf{b}+3\mathbf{c}=6\mathbf{c}=8\mathbf{a}=-8\mathbf{b}$$$$$$

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

$$\frac{ax+b}{\sqrt{1+3x}-2}=\frac{(ax+b)(\sqrt{1+3x}+2)}{x-1}=3c$$$$\sqrt{1+3x}=\frac{3c(x-1)}{ax+b}-2$$$$1+3x=4+\frac{9c^2{(x-1)}^2}{{(ax+b)}^2}-\frac{12c(x-1)(ax+b)}{{(ax+b)}^2}$$$$(1+3x){(ax+b)}^2=4{(ax+b)}^2-12c(x-1)(ax+b)+9c^2{(x-1)}^2$$$$3x{(ax+b)}^2=3{(ax+b)}^2-12c(ax+b)(x-1)+9c^2{(x-1)}^2$$$$3(x-1){(ax+b)}^2+12(x-1)(ax+b)-9c^2{(x-1)}^2=0$$$${(ax+b)}^2+4c(ax+b)-3c^2(x-1)$$$${(ax+b+2c)}^2-c^2(1+3x)=0$$$$c\ne 0$$$$4={\left(\frac{a+b+2c}{c}\right)}^2\Rightarrow \frac{\mathrm{a}+\mathrm{b}+2\mathrm{c}}{\mathrm{c}}=2$$$$$$$$\{\begin{array}{l}c=\mathbb{R}-\left\{0\right\}\\ a=-bet(a,b)\in {\mathbb{R}}^2-\{0,0\}\end{array}$$$$$$

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