\(\lim\limits_{x \rightarrow 2} \dfrac{(2x\:-\:1)^2\:-\:9}{x^2 \:-\:5x + 6} = \dotso\)
Answer: B
\(\lim\limits_{x \rightarrow 2} \dfrac{4x^2\:-\:4x\:-\:8}{x^2 \:-\:5x + 6}\)
\(\lim\limits_{x \rightarrow 2} \dfrac{(4x + 4 )\cancel{(x \:-\:2)}}{(x\:-\:3)\cancel{(x\:-\:2)}}\)
\(\lim\limits_{x \rightarrow 2} \dfrac{4x + 4}{x\:-\:3}\)
\(\dfrac{4(2) + 4}{2\:-\:3}\)
\(\dfrac{12}{-1}=-12\)
\(\lim\limits_{x \rightarrow 7} \dfrac{2x^2\:-\:9x\:-\:35}{3x^2 \:-\:23x + 14} = \dotso\)
Answer: A
Cara 1:
\(\lim\limits_{x \rightarrow 7} \dfrac{2x^2\:-\:9x\:-\:35}{3x^2 \:-\:23x + 14}\)
Karena substitusi \(x = 7\) menghasilkan\(\dfrac{0}{0}\) maka faktorkan pembilang dan penyebut.
\(\lim\limits_{x \rightarrow 7} \dfrac{(2x + 5)(x\:-\:7)}{(3x\:-\:2)(x\:-\:7)}\)
\(\lim\limits_{x \rightarrow 7} \dfrac{(2x + 5)\cancel{(x\:-\:7)}}{(3x\:-\:2)\cancel{(x\:-\:7)}}\)
\(\lim\limits_{x \rightarrow 7} \dfrac{2x + 5}{3x\:-\:2}\)
\(\dfrac{2(7) + 5}{3(7)\:-\:2}\)
\(\dfrac{19}{19}\)
\(1\)
Cara 2:
Karena substitusi\(x = 7\) menghasilkan\(\dfrac{0}{0}\) maka bisa digunakan metode L’hopital (pembilang dan penyebut diturunkan)
\(\lim\limits_{x \rightarrow 7} \dfrac{2x^2\:-\:9x\:-\:35}{3x^2 \:-\:23x + 14}\)
\(\lim\limits_{x \rightarrow 7} \dfrac{4x\:-\:9}{6x\:-\:23}\)
\(\dfrac{4(7)\:-\:9}{6(7)\:-\:23}\)
\(\dfrac{19}{19}\)
\(1\)
Answer: C
Cara 1:
Karena substitusi \(x = 3\) hasil limitnya \(\dfrac{0}{0}\) maka faktorkan pembilang dan penyebut
\(\color{blue} a^3\:-\:b^3 = (a\:-\:b)(a^2 + ab + b^2)\)
\(\lim\limits_{x \rightarrow 3} \dfrac{(x\:-\:3)(x^2 + 3x + 9)}{(x\:-\:3)(x\:-\:4)}\)
\(\lim\limits_{x \rightarrow 3} \dfrac{\cancel{(x\:-\:3)}(x^2 + 3x + 9)}{\cancel{(x\:-\:3)}(x\:-\:4)}\)
\(\lim\limits_{x \rightarrow 3} \dfrac{x^2 + 3x + 9}{x\:-\:4}\)
\(\dfrac{3^2 + 3(3) + 9}{3\:-\:4}\)
\(\dfrac{27}{-1}\)
\(-27\)
Cara 2:
Karena substitusi \(x = 3\) hasil limitnya \(\dfrac{0}{0}\) maka bisa digunakan metode L’hopital
\(\lim\limits_{x \rightarrow 3} \dfrac{x^3\:-\:27}{x^2\:-\:7x + 12}\)
\(\lim\limits_{x \rightarrow 3} \dfrac{3x^2}{2x\:-\:7}\)
\(\dfrac{3(3)^2}{2(3)\:-\:7}\)
\(\dfrac{27}{-1}\)
\(-27\)
\(\lim\limits_{x \rightarrow 5} \dfrac{(x^2\:-\:25)}{\sqrt{x^2 + 24}\:-\:7} = \dotso\)
Answer: D
\(\lim\limits_{x \rightarrow 5} \dfrac{(x^2 \:-\:25)}{\sqrt{x^2 + 24}\:-\:7} \times \color{red} \dfrac{\sqrt{x^2 + 24}+7}{\sqrt{x^2 + 24}+7}\)
\(\lim\limits_{x \rightarrow 5}\dfrac{(x^2 \:-\:25)(\sqrt{x^2 + 24}+7)}{x^2 + 24\:-\:49}\)
\(\lim\limits_{x \rightarrow 5}\dfrac{\cancel{(x^2 \:-\:25)}(\sqrt{x^2 + 24}+7)}{\cancel{x^2 \:-\:25}}\)
\(\lim\limits_{x \rightarrow 5} (\sqrt{x^2 + 24}+7)\)
\(\sqrt{5^2 + 24}+7\)
\(7 + 7\)
\(14\)
\(\lim\limits_{x \rightarrow 0} \dfrac{\sqrt{x^2 + 9x}}{x\sqrt{x} + \sqrt{x}} = \dotso\)
Answer: C
\(\lim\limits_{x \rightarrow 0} \dfrac{\sqrt{x(x + 9)}}{\sqrt{x}(x + 1)}\)
\(\lim\limits_{x \rightarrow 0} \dfrac{\cancel{\sqrt{x}}\sqrt{x + 9}}{\cancel{\sqrt{x}}(x + 1)}\)
\(\dfrac{\sqrt{0 + 9}}{0 + 1}\)
\(3\)
\(\lim\limits_{x \rightarrow 1} \dfrac{(\sqrt{5\:-\:x}\:-\:2)(\sqrt{2\:-\:x} + 1)}{1\:-\:x} = \dotso\)
Answer: E
\(\lim\limits_{x \rightarrow 1} \dfrac{(\sqrt{5\:-\:x}\:-\:2)(\sqrt{2\:-\:x} + 1)}{1\:-\:x}\times \color{red} \dfrac{\sqrt{5\:-\:x} + 2}{\sqrt{5\:-\:x} + 2}\)
\(\lim\limits_{x \rightarrow 1} \dfrac{\left[(\sqrt{5\:-\:x})^2\:-\:2^2\right](\sqrt{2\:-\:x} + 1)}{(1\:-\:x)(\sqrt{5\:-\:x} + 2)}\)
\(\lim\limits_{x \rightarrow 1} \dfrac{\cancel{(1\:-\:x)}(\sqrt{2\:-\:x} + 1)}{\cancel{(1\:-\:x)}(\sqrt{5\:-\:x} + 2)}\)
\(\dfrac{\sqrt{2\:-\:1} + 1}{\sqrt{5\:-\:1} + 2}\)
\(\dfrac{2}{4}\)
\(\dfrac{1}{2}\)
\(\lim\limits_{x \rightarrow 9} \dfrac{(\sqrt{x}\:-\:3)(\sqrt{x} + 9)}{x\:-\:9} = \dotso\)
Answer: B
\(\lim\limits_{x \rightarrow 9} \dfrac{(\sqrt{x}\:-\:3)(\sqrt{x} + 9)}{(\sqrt{x})^2\:-\:3^2}\)
\(\lim\limits_{x \rightarrow 9} \dfrac{\cancel{(\sqrt{x}\:-\:3)}(\sqrt{x} + 9)}{(\sqrt{x} + 3)\cancel{(\sqrt{x}\:-\:3)}}\)
\(\lim\limits_{x \rightarrow 9} \dfrac{\sqrt{x} + 9}{\sqrt{x} + 3}\)
\(\dfrac{\sqrt{9} + 9}{\sqrt{9} + 3}\)
\(\dfrac{12}{6}\)
\(2\)
Jika \(f(x) = 2x^2\:-\:1\), maka nilai dari \(\lim\limits_{h \rightarrow 0} \dfrac{f(3 + h) \:-\:f(3)}{2h} = \dotso\)
Answer: C
\(f(3 + h) = 2(3 + h)^2\:-\:1\)
\(f(3 + h) = 2(9 + 6h + h^2)\:-\:1\)
\(f(3 + h) = 18 + 12h + h^2\:-\:1\)
\(f(3 + h) = h^2 + 12h + 17\)
\(f(3) = 2(3)^2\:-\:1\)
\(f(3) = 18\:-\:1 = 17\)
\(\lim\limits_{h \rightarrow 0} \dfrac{f(3 + h) \:-\:f(3)}{2h} = \lim\limits_{h \rightarrow 0} \dfrac{h^2 + 12h + 17\:-\:17}{2h}\)
\lim\limits_{h \rightarrow 0} \dfrac{\cancel{h}(h + 12)}{2\cancel{h}}\)
\(\lim\limits_{h \rightarrow 0} \dfrac{h + 12}{2}\)
\(\dfrac{0 + 12}{2} = 6\)
Answer: E
Gunakan rumus:
\(\color{blue} (x\:-\:a) = (\sqrt[3] {x} \:-\:\sqrt[3] {a} )(\sqrt[3] {x^2} + \sqrt[3] {x}\cdot \sqrt[3] {a} + \sqrt[3] {a^2})\)
\(\color{blue} (x\:-\:8) = (\sqrt[3] {x} \:-\:\sqrt[3] {8} )(\sqrt[3] {x^2} + \sqrt[3] {x}\cdot \sqrt[3] {8} + \sqrt[3] {8^2})\)
\((x\:-\:8) = (\sqrt[3] {x} \:-\:2)(\sqrt[3] {x^2} + 2\sqrt[3] {x} + 4)\)
Jadi bentuk sekawan untuk \(\sqrt[3] {x} \:-\:2\) adalah \(\color{red} \sqrt[3] {x^2} + 2\sqrt[3] {x} + 4\)
\(\lim\limits_{x \rightarrow 8} \dfrac{x\:-\:8}{\sqrt[3] {x} \:-\:2} \times \color{red} \dfrac{\sqrt[3] {x^2} + 2\sqrt[3] {x} + 4}{\sqrt[3] {x^2} + 2\sqrt[3] {x} + 4}\)
\(\lim\limits_{x \rightarrow 8} \dfrac{\cancel{(x\:-\:8)}(\sqrt[3] {x^2} + 2\sqrt[3] {x} + 4)}{\cancel{x\:-\:8}}\)
\(\lim\limits_{x \rightarrow 8} \sqrt[3] {x^2} + 2\sqrt[3] {x} + 4\)
\(\sqrt[3] {8^2} + 2\sqrt[3] {8} + 4\)
\(4 + 2\cdot 2 + 4\)
\(12\)
\(\lim\limits_{x \rightarrow 2}\left( \dfrac{2}{x^2\:-\:4}\:-\:\dfrac{3}{x^2 + 2x\:-\:8}\right) = \dotso\)
Answer: D
\(\lim\limits_{x \rightarrow 2}\left( \dfrac{2}{(x + 2)(x\:-\:2)}\:-\:\dfrac{3}{(x + 4)(x\:-\:2)}\right)\)
\(\lim\limits_{x \rightarrow 2} \dfrac{2(x + 4)\:-\:3(x + 2)}{(x + 2)(x\:-\:2)(x + 4)}\)
\(\lim\limits_{x \rightarrow 2} \dfrac{2x + 8\:-\:3x\:-\:6}{(x + 2)(x\:-\:2)(x + 4)}\)
\(\lim\limits_{x \rightarrow 2} \dfrac{(-x + 2)}{(x + 2)(x\:-\:2)(x + 4)}\)
\(\lim\limits_{x \rightarrow 2} \dfrac{-\cancel{(x\:-\:2)}}{(x + 2)\cancel{(x\:-\:2)}(x + 4)}\)
\(\lim\limits_{x \rightarrow 2} \dfrac{-1}{(x + 2)(x + 4)}\)
\(\dfrac{-1}{(4)(6)}\)
\(-\dfrac{1}{24}\)
