Cho hàm số g(x) = \(\frac{\sqrt{2x\:+\:2}-\:\sqrt{3x\:+\:1}}{mx^2\:-\:m}\)với m khác 0 và f(x) = \(\frac{8x^{2016}\:-\:24x^{2015}}{x^{2017}\:+\:2x^{2016}\:-\:15x^{2015}}\). Ta có: lim g(x) khi x -> 1 = lim f(x) khi x -> 3. Lúc đó giá trị tham số m bằng:
A. \(\frac{-1}{64}\)
B. \(\frac{-1}{8}\)
C. 8
D. \(\frac{1}{64}\)
\(\lim\limits_{x\rightarrow3}f\left(x\right)=\lim\limits_{x\rightarrow3}\frac{8x^{2016}-24x^{2015}}{x^{2017}+2x^{2016}-15x^{2015}}=\lim\limits_{x\rightarrow3}\frac{8\left(x-3\right)}{x^2+2x-15}=\lim\limits_{x\rightarrow3}\frac{8\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}=\lim\limits_{x\rightarrow3}\frac{8}{x+5}=1\)
\(\lim\limits_{x\rightarrow1}g\left(x\right)=\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+2}-2+2-\sqrt{3x+1}}{m\left(x-1\right)\left(x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{2\left(x-1\right)}{\sqrt{2x+2}+2}-\frac{3\left(x-1\right)}{2+\sqrt{3x+1}}}{m\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{2}{\sqrt{2x+2}+2}-\frac{3}{2+\sqrt{3x+1}}}{m\left(x+1\right)}=\frac{\frac{2}{4}-\frac{3}{4}}{2m}=-\frac{1}{8m}\)
\(\Rightarrow-\frac{1}{8m}=1\Rightarrow m=-\frac{1}{8}\)
