PAS MATEMATIKA BLOG: Contoh Soal 2 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{ll} 6. & Nilai yang memenuhi \\ lim_{x \to \infty} (\sqrt{8 x - 2020} - \sqrt{4 x + 2021}) adalah... . \\ \begin{array}{llll} a . & - \infty \\ b . & 0 \\ c . & 1 \\ d . & 2 \\ e . & \infty \end{array} \\ Jawab : e \\ \begin{aligned} lim_{x \to \infty} (\sqrt{8 x - 2020} - \sqrt{4 x + 2021}) \\ = lim_{x \to \infty} (\sqrt{8 x - 2020} - \sqrt{4 x + 2021}) \times \frac{\sqrt{8 x - 2020} + \sqrt{4 x + 2021}}{\sqrt{8 x - 2020} + \sqrt{4 x + 2021}} \\ = lim_{x \to \infty} \frac{(8 x - 2020) - (4 x + 2021)}{\sqrt{8 x - 2020} + \sqrt{4 x + 2021}} \\ = lim_{x \to \infty} \frac{4 x - 4041}{\sqrt{8 x - 2020} + \sqrt{4 x + 2021}} \times \frac{\frac{1}{x}}{\frac{1}{x}} \\ = lim_{x \to \infty} \frac{4 - \frac{4041}{x}}{\frac{1}{x} (\sqrt{8 x - 2020} + \sqrt{4 x + 2021})} \\ = lim_{x \to \infty} \frac{4 - \frac{4041}{x}}{(\sqrt{\frac{8 x}{x^{2}} - \frac{2020}{x^{2}}} + \sqrt{\frac{4 x}{x^{2}} + \frac{2021}{x^{2}}})} \\ = lim_{x \to \infty} \frac{4 - \frac{4041}{x}}{(\sqrt{\frac{8}{x} - \frac{2020}{x}} + \sqrt{\frac{4}{x} + \frac{2021}{x}})} \\ = \frac{4 - 0}{\sqrt{0 - 0} + \sqrt{0 + 0}} \\ = \frac{4}{0} \\ = \infty \end{aligned} \end{array}$

$\begin{array}{l} 7. & Nilai yang memenuhi \\ lim_{x \to \infty} (\sqrt{8 x - 2020} + \sqrt{4 x + 2021}) adalah... . \\ \begin{array}{llll} a . & - \infty \\ b . & 0 \\ c . & 1 \\ d . & 2 \\ e . & \infty \end{array} \\ Jawab : e \\ lim_{x \to \infty} (\sqrt{8 x - 2020} + \sqrt{4 x + 2021}) \\ = \sqrt{\infty} + \sqrt{\infty} \\ = \infty \end{array}$

$\begin{array}{ll} 8. & Nilai yang memenuhi \\ lim_{x \to \infty} (\sqrt{4 x - 2020} - \sqrt{8 x + 2021}) adalah... . \\ \begin{array}{llll} a . & - \infty \\ b . & 0 \\ c . & 1 \\ d . & 2 \\ e . & \infty \end{array} \\ Jawab : a \\ \begin{aligned} lim_{x \to \infty} (\sqrt{4 x - 2020} - \sqrt{8 x + 2021}) \\ = lim_{x \to \infty} (\sqrt{4 x - 2020} - \sqrt{8 x + 2021}) \times \frac{\sqrt{4 x - 2020} + \sqrt{8 x + 2021}}{\sqrt{4 x - 2020} + \sqrt{8 x + 2021}} \\ = lim_{x \to \infty} \frac{(4 x - 2020) - (8 x + 2021)}{\sqrt{4 x - 2020} + \sqrt{8 x + 2021}} \\ = lim_{x \to \infty} \frac{- 4 x - 4041}{\sqrt{4 x - 2020} + \sqrt{8 x + 2021}} \times \frac{\frac{1}{x}}{\frac{1}{x}} \\ = lim_{x \to \infty} \frac{- 4 - \frac{4041}{x}}{\frac{1}{x} (\sqrt{4 x - 2020} + \sqrt{8 x + 2021})} \\ = lim_{x \to \infty} \frac{- 4 - \frac{4041}{x}}{(\sqrt{\frac{4 x}{x^{2}} - \frac{2020}{x^{2}}} + \sqrt{\frac{8 x}{x^{2}} + \frac{2021}{x^{2}}})} \\ = lim_{x \to \infty} \frac{- 4 - \frac{4041}{x}}{(\sqrt{\frac{4}{x} - \frac{2020}{x}} + \sqrt{\frac{8}{x} + \frac{2021}{x}})} \\ = \frac{- 4 - 0}{\sqrt{0 - 0} + \sqrt{0 + 0}} \\ = \frac{- 4}{0} \\ = - \infty \end{aligned} \end{array}$

$\begin{array}{ll} 9. & Nilai \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} = . . . . \\ \begin{array}{lll} a . - 1 \\ b . 1 \\ c . 2 \\ d . 4 \\ e . 8 \end{array} \\ Jawab : c \\ \begin{aligned} \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} \\ = \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} \times \frac{\sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x}}{\sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x}} \\ = \underset{x \to \infty}{Lim} \frac{4 x^{2} + 3 x - (4 x^{2} - 5 x)}{\sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x}} \\ = \underset{x \to \infty}{Lim} \frac{3 x + 5 x}{\sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x}} \times \frac{(\frac{1}{x})}{(\sqrt{\frac{1}{x^{2}}})} \\ = \frac{3 + 5}{\sqrt{4} + \sqrt{4}} \\ = \frac{8}{4} \\ = 2 \end{aligned} \end{array}$

$\begin{aligned} ada cara lain yang lebih sede & rhana, yaitu: \\ . \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} & = \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} \\ {\begin{cases} a & = 4 \\ b & = 3 \\ p & = - 4 \end{cases} \\ Jika \\ \underset{x \to \infty}{Lim} \sqrt{a x^{2} + b x + c} & - \sqrt{a x^{2} + p x + q} = \frac{b - p}{2 \sqrt{a}} \\ Sehingga \\ \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} & = \frac{3 - (- 5)}{2 \sqrt{4}} \\ = \frac{8}{2.2} \\ = 2 \end{aligned}$

$\begin{array}{ll} 10. & Nilai \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x} = . . . . \\ \begin{array}{lll} a . \infty \\ b . 1 \\ c . 2 \\ d . 4 \\ e . 8 \end{array} \\ Jawab : a \\ \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x} \\ = \sqrt{\infty} + \sqrt{\infty} = \infty \end{array}$

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$\begin{aligned} Sebagai & CATATAN di sini \\ Sifat-sif & at bilangan tak hingga \\ (1) & \infty + \infty = \infty \\ (2) & - \infty + (- \infty) = - \infty \\ (3) & \infty \times \infty = \infty \\ (4) & - \infty + (- \infty) = \infty \\ (5) & k . \infty = \infty, k positif \\ (6) & k . (- \infty) = - \infty, k positif \\ (7) & k . \infty = - \infty, k negatif \\ (8) & k . (- \infty) = \infty, k negatif \\ yang ha & rus dihindari \\ (1) & \infty - \infty, bentuk tak tentu \\ (2) & \frac{\infty}{\infty}, - \frac{\infty}{\infty}, dan \frac{0}{0} \end{aligned}$

PAS MATEMATIKA BLOG

Contoh Soal 2 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

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