Materi : Limit Fungsi Kelas / Semester : 11 / 2 [tex]\lim_{(y-2) \to \ 0} \frac{1}{y-2} ( \frac{1}{2y^{2}-y-3} - \frac{2}{y^{2}+y} )[/tex]
Matematika
rikuta7
Pertanyaan
Materi : Limit Fungsi
Kelas / Semester : 11 / 2
[tex]\lim_{(y-2) \to \ 0} \frac{1}{y-2} ( \frac{1}{2y^{2}-y-3} - \frac{2}{y^{2}+y} )[/tex]
Kelas / Semester : 11 / 2
[tex]\lim_{(y-2) \to \ 0} \frac{1}{y-2} ( \frac{1}{2y^{2}-y-3} - \frac{2}{y^{2}+y} )[/tex]
1 Jawaban
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1. Jawaban david993
[tex]lim(y - 2) \frac{1}{y - 2} ( \frac{1}{2 {y}^{2} - y - 3} - \frac{2}{ {y}^{2} + y } ) \\ lim(y - 2) \frac{1}{y - 2} ( \frac{ {y}^{2} + y - 2(2 {y}^{2} - y - 3) }{(2 {y}^{2} - y - 3)( {y}^{2} + y) } ) \\lim(y - 2) \frac{ - 3 {y}^{2} + 3y + 6 }{(y - 2)( {y}^{2} + y)(2 {y}^{2} - y - 3) } \\ lim(y - 2) \frac{ - 3(y - 2)(y + 1)}{(y - 2)(y)(y + 1)(2 {y}^{2} - y - 3) } \\ lim(y - 2) \frac{ - 3}{y(2 {y}^{2} - y - 3)} \\ \frac{ - 3}{2(2 ( {2}^{2}) - (2) - 3 )} \\ = - \frac{3}{2 \times 3 } = - \frac{1}{2} [/tex]