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[Гост]

лим. х->-1/2 (6x^2+5x+3)/(4x^2-1) + 1/(2x+1)
Получавам (-1/24) , а в отговора пише (-1/4)
Дали вярното е някъде там ?

[ammornil]

лим. х->-1/2 (6x^2+5x+3)/(4x^2-1) + 1/(2x+1)
Получавам (-1/24) , а в отговора пише (-1/4)
Дали вярното е някъде там ?

[tex]\lim_{x \to -\frac{1}{2}} \left( \underbrace{ \frac{6x^{2}+5x+3}{(2x-1)(2x+1)}+\frac{1}{2x+1}}_{(2x-1)(2x+1)} \right) = \lim_{x \to -\frac{1}{2}} \left( \frac{6x^{2}+5x+3+2x-1}{(2x-1)(2x+1)} \right) = \lim_{x \to -\frac{1}{2}} \frac{6x^{2}+7x+2}{4\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)}[/tex]

[tex]6x^{2}+7x+2 = 0 \rightarrow x_{1,2}=\frac{-7 \mp \sqrt{7^{2}-4.6.2}}{2.6} \Rightarrow \begin{cases}x_{1}=\frac{-7-1}{12}=-\frac{2}{3} \\ x_{2}=\frac{-7+1}{12}=-\frac{1}{2} \end{cases} \Rightarrow 6x^{2}+7x+2 = 6\left(x+\frac{2}{3}\right)\left(x+\frac{1}{2}\right)[/tex]

[tex]\Rightarrow \lim_{x \to -\frac{1}{2}} \frac{6x^{2}+7x+2}{4\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)}=\lim_{x \to -\frac{1}{2}} \frac{6\left(x+\frac{2}{3}\right)\left(x+\frac{1}{2}\right)}{4\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)}= \lim_{x \to -\frac{1}{2}} \frac{3\left(x+\frac{2}{3}\right)}{2\left(x-\frac{1}{2}\right)}=...=\frac{3\left(\underbrace{-\frac{1}{2}+\frac{2}{3}}_{6}\right)}{2\left(-\frac{1}{2}-\frac{1}{2}\right)}=[/tex]

[tex]=\frac{3.\large{\frac{3.(-1)+2.2}{6}}}{2.(-1)}=\frac{\frac{1}{2}}{-2}=-\frac{1}{4}[/tex]