[tex] \begin{aligned} \lim_{x\to\infty} \frac{ 5x-2}{2x^2-5x+8 } &=\lim_{x\to\infty} \frac{ 5x-2}{2x^2-5x+8 } \cdot \frac{ \frac{ 1}{x^2}}{ \frac{1}{x^2}}\\ &= \lim_{x\to\infty} \frac{\frac5x- \frac{ 2}{x^2}}{2-\frac5x+ \frac{ 8}{x^2}} \\ &= \frac{ 0-0}{2-0+0 }\\ &= \frac{0}{2} \\ &= 0\end{aligned} [/tex]
[tex] \begin{aligned} f(x) &= (x-17)(1-3x^2) \\ &\downarrow \\ f'(x) &= 1(1-3x^2)+(x-17)(-6) \\ f'(x) &= 1-3x^2-6x+ 102 \\ f'(x) &= -3x^2-6x+103\end{aligned} [/tex]
[tex]\begin{aligned} f(x) &= \frac{ 4-11x}{ 3x+7}, \quad x \ne - \frac{7}{3} \\ &\downarrow \\f'(x) &= \frac{-11(3x+7)-(4-11x)(3) }{(3x+7)^2 }\\ f'(x) &= \frac{ -33x-77-12+33x}{ (3x+7)^2}\\ f'(x) &=- \frac{ 89}{(3x+7)^2 } ,\quad x \ne - \frac{7}{3}\end{aligned} [/tex]
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Verified answer
Soal 10
[tex] \begin{aligned} \lim_{x\to\infty} \frac{ 5x-2}{2x^2-5x+8 } &=\lim_{x\to\infty} \frac{ 5x-2}{2x^2-5x+8 } \cdot \frac{ \frac{ 1}{x^2}}{ \frac{1}{x^2}}\\ &= \lim_{x\to\infty} \frac{\frac5x- \frac{ 2}{x^2}}{2-\frac5x+ \frac{ 8}{x^2}} \\ &= \frac{ 0-0}{2-0+0 }\\ &= \frac{0}{2} \\ &= 0\end{aligned} [/tex]
Soal 11
[tex] \begin{aligned} f(x) &= (x-17)(1-3x^2) \\ &\downarrow \\ f'(x) &= 1(1-3x^2)+(x-17)(-6) \\ f'(x) &= 1-3x^2-6x+ 102 \\ f'(x) &= -3x^2-6x+103\end{aligned} [/tex]
Soal 12
[tex]\begin{aligned} f(x) &= \frac{ 4-11x}{ 3x+7}, \quad x \ne - \frac{7}{3} \\ &\downarrow \\f'(x) &= \frac{-11(3x+7)-(4-11x)(3) }{(3x+7)^2 }\\ f'(x) &= \frac{ -33x-77-12+33x}{ (3x+7)^2}\\ f'(x) &=- \frac{ 89}{(3x+7)^2 } ,\quad x \ne - \frac{7}{3}\end{aligned} [/tex]