\(\color{blue} \lim\limits_{x \rightarrow \infty} \dfrac{ax^m + bx +c}{px^n + qx + r} = \dfrac{a}{p} \text{ jika } m = n\)
\(\color{blue} \lim\limits_{x \rightarrow \infty} \dfrac{ax^m + bx +c}{px^n + qx + r} = 0 \text{ jika } m < n\)
\(\color{blue} \lim\limits_{x \rightarrow \infty} \dfrac{ax^m + bx +c}{px^n + qx + r} = \infty \text{ jika } m > n\)
\(\color{blue} \lim\limits_{x \rightarrow \infty} \sqrt{ax^2 + bx + c}\:-\:\sqrt{px^2 + qx + r} = \dfrac{b\:-\:q}{2\sqrt{a}}, \text{ jika } a = p\)
\(\color{blue} \lim\limits_{x \rightarrow \infty} \sqrt [n] {ax^n+ bx^{n\:-\:1} + cx^{n\:-\:2} + \dotso}\:-\:\sqrt [n] {px^n + qx^{n\:-\:1} + rx^{n\:-\:2} + \dotso} = \dfrac{b\:-\:q}{n\cdot\sqrt [n] {a^{n\:-\:1}}}, \text{ jika } a = p\)
\(\lim\limits_{x \rightarrow \infty} \dfrac{2\:-\:6x^5 + 2x^2}{2x^5 + x^3 + 1} = \dotso\)
Answer: A
\(\lim\limits_{x \rightarrow \infty} \dfrac{2\:-\:6x^5 + 2x^2}{2x^5 + x^3 + 1} = \lim\limits_{x \rightarrow \infty} \dfrac{-6x^5 + \dotso}{2x^5 + \dotso}\)
Bagi pembilang dan penyebut dengan \(x\) pangkat tertinggi yaitu \(x^5\)
\(\lim\limits_{x \rightarrow \infty} \dfrac{-6\cancel{x^5} + \dotso}{2\cancel{x^5} + \dotso} = \dfrac{-6}{2} = -3\)
\(\lim\limits_{x \rightarrow \infty} \dfrac{8x + 5 \:-\:\sqrt{4x^2 + 2x + 5}}{7\:-\:4x + \sqrt{25x^2 \:-\:1}} = \dotso\)
Answer: E
\(\lim\limits_{x \rightarrow \infty}\dfrac{8x + 5 \:-\:\sqrt{4x^2 + 2x + 5}}{7\:-\:4x + \sqrt{25x^2 \:-\:1}} \)
Fokus pada suku dengan \(x\) pangkat tertinggi saja
\(\lim\limits_{x \rightarrow \infty}\dfrac{8x\:-\:\sqrt{4x^2 + \dotso}}{-4x + \sqrt{25x^2\dotso}} \)
Bagi pembilang dan penyebut dengan \(x\) pangkat tertinggi yaitu \(x^1\)
\(\lim\limits_{x \rightarrow \infty}\dfrac{8\:-\:2}{-4 + 5}\)
\(\dfrac{6}{1} = 6\)
\(\lim\limits_{n \rightarrow \infty} \dfrac{12n^2 [2 + 6 + 10 + 14 + \dotso + (4n \:-\:2)]}{6[-6\:-\:3 + 0 + 3 + 6 + \dotso (3n\:-\:9)]^2}= 9k\), nilai \(k\) yang memenuhi adalah…
Answer: E
Rumus jumlah n suku pertama deret aritmetika:
\(\color{blue} S_n = \dfrac{n}{2} (a + U_n)\)
\(\lim\limits_{n \rightarrow \infty} \dfrac{12n^2 \left[\dfrac{n}{2}(2 + 4n\:-\:2)\right]}{6\left[\dfrac{n}{2}(-6 + 3n\:-\:9)\right]^2}\)
\(\lim\limits_{n \rightarrow \infty} \dfrac{12n^2 \left[\dfrac{n}{2}(4n)\right]}{6\left[\dfrac{n}{2}(3n\:-\:15)\right]^2}\)
\(\lim\limits_{n \rightarrow \infty} \dfrac{12n^2 \cdot 2n^2}{6\cdot \dfrac{n^2}{4}(3n\:-\:15)^2}\)
\(\lim\limits_{n \rightarrow \infty} \dfrac{24n^4}{\dfrac{3n^2}{2}(3n\:-\:15)^2}\)
\(\lim\limits_{n \rightarrow \infty} \dfrac{24n^4}{\dfrac{3}{2}n^2 \cdot (9n^2 + \dotso)}\)
\(\lim\limits_{n \rightarrow \infty} \dfrac{24n^4}{\dfrac{27}{2}n^4 + \dotso}\)
bagi pembilang dan penyebut dengan \(n^4\)
\(\dfrac{24}{\dfrac{27}{2}}\)
\(\dfrac{16}{9}\)
\(\dfrac{16}{9} = 9k \rightarrow k = 16\)
\(\lim\limits_{x \rightarrow \infty} \dfrac{2^{x+2} + 3^{2x + 1} + 4^{x + 3}}{2^{2x + 1} + 5^{x + 3} + 9^{x\:-\:1}} = \dotso\)
Answer: B
\(\lim\limits_{x \rightarrow \infty} \dfrac{2^x \cdot 2^2 + 3^{2x} \cdot 3^1 + 4^x \cdot 4^3}{2^{2x} \cdot 2^1 + 5^x \cdot 5^3 + 9^x \cdot 9^{-1}}\)
\(\lim\limits_{x \rightarrow \infty} \dfrac{4\cdot 2^x + 3\cdot 3^{2x} + 64 \cdot 2^{2x}}{2 \cdot 2^{2x} + 125 \cdot 5^x + \dfrac{1}{9} \cdot 3^{2x}}\)
Bagi pembilang dan penyebut dengan \(3^{2x}\)
\(\dfrac{3}{\dfrac{1}{9}}\)
\(27\)
\(\lim\limits_{x\rightarrow \infty} \dfrac{(2x + 1)^2 (4x + 3)(3x\:-\:9)(6x\:-\:11)}{(2x + 1)^6 \:-\:(2x\:-\:3)^6} = \dotso\)
Answer: D
\(\lim\limits_{x \rightarrow \infty} \dfrac{(2x)^2 (4x) (3x) (6x)}{(\cancel{(2x)^6} + 6(2x)^5 \cdot 1+ \dotso)\:-\:(\cancel{(2x)^6}\:-\:6(2x)^5\cdot 3 + \dotso)}\)
\(\lim\limits_{x \rightarrow \infty} \dfrac{288x^5 + \dotso}{192x^5 + 576x^5 + \dotso}\)
\(\dfrac{288}{768}\)
\(\dfrac{3}{8}\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{4x^2 + 8x \:-\:5} \:-\:\sqrt{4x^2 \:-\:4x + 10} = \dotso\)
Answer: C
\(\color{blue} \lim\limits_{x \rightarrow \infty} \sqrt{ax^2 + bx + c}\:-\:\sqrt{px^2 + qx + r} = \dfrac{b\:-\:q}{2\sqrt{a}}, \text{ jika } a = p\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{4x^2 + 8x \:-\:5} \:-\:\sqrt{4x^2 \:-\:4x + 10} =\color{blue} \dfrac{b\:-\:q}{2\sqrt{a}}\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{4x^2 + 8x \:-\:5} \:-\:\sqrt{4x^2 \:-\:4x + 10} =\color{blue} \dfrac{8\:-\:(-4)}{2\sqrt{4}}\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{4x^2 + 8x \:-\:5} \:-\:\sqrt{4x^2 \:-\:4x + 10} =\color{blue} \dfrac{12}{4} = 3\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{9x^2\:-\:12x + 7} \:-\:\sqrt{9x^2 + 24x\:-\:11} = \dotso\)
Answer: C
\(\lim\limits_{x \rightarrow \infty} \sqrt{9x^2\:-\:12x + 7} \:-\:\sqrt{9x^2 + 24x\:-\:11} =\color{blue} \dfrac{b\:-\:q}{2\sqrt{a}}\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{9x^2\:-\:12x + 7} \:-\:\sqrt{9x^2 + 24x\:-\:11} =\color{blue} \dfrac{-12\:-\:24}{2\sqrt{9}}\)
\(\lim\limits_{x \rightarrow \infty}\sqrt{9x^2\:-\:12x + 7} \:-\:\sqrt{9x^2 + 24x\:-\:11} =\color{blue} \dfrac{-36}{6} = -6\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2\:-\:2x + 5} \:-\:x\:-\:3 = \dotso\)
Answer: A
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2\:-\:2x + 5} \:-(x + 3)\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2\:-\:2x + 5} \:-\sqrt{(x + 3)^2}\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2\:-\:2x + 5} \:-\sqrt{x^2 + 6x + 9}\)
\(\dfrac{-2\:-\:6}{2\sqrt{1}}\)
\(\dfrac{-8}{2}\)
\(-4\)
\(\lim\limits_{x \rightarrow \infty} \left(\sqrt{x^2 + x} \:-\:\sqrt{9x^2\:-\:6x} + \sqrt{4x^2 \:-\:12x}\right) = \dotso\)
Answer: A
\(\lim\limits_{x \rightarrow \infty} \left(\sqrt{x^2 + x} \:-\:\sqrt{9x^2\:-\:6x} + \sqrt{4(x^2\:-\:3x)}\right)\)
\(\lim\limits_{x \rightarrow \infty} \left(\sqrt{x^2 + x} \:-\:\sqrt{9x^2\:-\:6x} + \color{blue} 2\sqrt{x^2\:-\:3x}\right)\)
\(\lim\limits_{x \rightarrow \infty} \left(\sqrt{x^2 + x} \:-\:\sqrt{9x^2\:-\:6x} + \color{red} 3\sqrt{x^2\:-\:3x}\:-\:\color{purple} 1\sqrt{x^2\:-\:3x}\right)\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2 + x} \:-\:\color{purple}\sqrt{x^2\:-\:3x}\color{black} + \lim\limits_{n \rightarrow \infty} \color{red}\sqrt{9x^2 \:-\:27x} \:-\:\color{black} \sqrt{9x^2\:-\:6x}\)
\(\dfrac{1\:-\:(-3)}{2\sqrt{1}} + \dfrac{-27\:-\:(-6)}{2\sqrt{9}}\)
\(2\:-\:\dfrac{21}{6}\)
\(-\dfrac{3}{2} = -1,5\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2 + 4x \:-\:1}\:-\:\sqrt{36x^2 + 6x + 5} + \sqrt{25x^2 + 10x + 1} =\dotso\)
Answer: C
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2 + 4x \:-\:1}\:-\:\sqrt{36x^2 + 6x + 5} + \sqrt{(5x + 1)^2}\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2 + 4x \:-\:1}\:-\:\sqrt{36x^2 + 6x + 5} + \color{blue} 5x + 1\)
\(\lim\limits_{x \rightarrow \infty} \sqrt{x^2 + 4x \:-\:1}\:-\:\sqrt{36x^2 + 6x + 5} + \color{red} 6x\:\color{purple} -\:x + 1\)
\(\lim\limits_{x \rightarrow \infty}\sqrt{x^2 + 4x \:-\:1}\:-\:(x\:-\:1) + \lim\limits_{n \rightarrow \infty} 6x \:-\: \sqrt{36x^2 + 6x + 5}\)
\(\lim\limits_{x \rightarrow \infty}\sqrt{x^2 + 4x \:-\:1}\:-\:\sqrt{(x\:-\:1)^2} + \lim\limits_{n \rightarrow \infty} \sqrt{(6x)^2} \:-\: \sqrt{36x^2 + 6x + 5}\)
\(\lim\limits_{x \rightarrow \infty}\sqrt{x^2 + 4x \:-\:1}\:-\:\sqrt{x^2\:-\:2x + 1} + \lim\limits_{n \rightarrow \infty} \sqrt{36x^2} \:-\: \sqrt{36x^2 + 6x + 5}\)
\(\dfrac{4\:-\:(-2)}{2\sqrt{1}} + \dfrac{0\:-\:6}{2\sqrt{36}}\)
\(3 \:-\: \dfrac{1}{2}\)
\(2,5\)
\(\lim\limits_{x \rightarrow \infty}\sqrt[3] {64x^3\:-\:12x^2 + 5x\:-\:12}\:-\:\sqrt[3] {64x^3 \:-\:4x^2 \:-\:3x\:-\:5} = \dotso\)
Answer: E
\(\lim\limits_{x \rightarrow \infty}\sqrt[3] {64x^3 \:-\: 12x^2 + 5x\:-\: 12}\:-\: \sqrt[3] {64x^3 \:-\:4x^2 \:-\:3x\:-\:5}\)
Rumus:
\(\color{blue} \lim\limits_{x \rightarrow \infty} \sqrt [n] {ax^n+ bx^{n\:-\:1} + cx^{n\:-\:2} + \dotso}\:-\:\sqrt [n] {px^n + qx^{n\:-\:1} + rx^{n\:-\:2} + \dotso} = \dfrac{b\:-\:q}{n\cdot\sqrt [n] {a^{n\:-\:1}}}, \text{ jika } a = p\)
\(n = 3 \text{ (akar pangkat 3)}\)
\(a = p = 64\)
\(b = -12 \text{ (koef. x² polinomial pertama)}\)
\(q = -4 \text{ (koef. x² polinomial kedua)}\)
\(\lim\limits_{x \rightarrow \infty}\sqrt [3] {64x^3\:-\:12x^2 + 5x\:-\:12}\:-\:\sqrt [3] {64x^3 \:-\:4x^2 \:-\:3x\:-\:5} = \dfrac{-12\:-\:(-4)}{3 \cdot \sqrt[3] {64^{3\:-\:1}}}\)
\(\lim\limits_{x \rightarrow \infty}\sqrt [3] {64x^3\:-\:12x^2 + 5x\:-\:12}\:-\:\sqrt [3] {64x^3 \:-\:4x^2 \:-\:3x\:-\:5} = \dfrac{-8}{3 \cdot \sqrt[3]{4^{6}}}\)
\(\lim\limits_{x \rightarrow \infty}\sqrt [3] {64x^3\:-\:12x^2 + 5x\:-\:12}\:-\:\sqrt [3] {64x^3 \:-\:4x^2 \:-\:3x\:-\:5} = \dfrac{-8}{3 \cdot 4^2}\)
\(\lim\limits_{n \rightarrow \infty}\sqrt [3] {64x^3\:-\:12x^2 + 5x\:-\:12}\:-\:\sqrt [3] {64x^3 \:-\:4x^2 \:-\:3x\:-\:5} = \dfrac{-8}{48} = -\dfrac{1}{6}\)
