Pada artikel ini kita akan mempelajari cara menghitung nilai limit dari fungsi-fungsi aljabar, khususnya polinom dan rasional. Adapun metode yang dipakai yakni substitusi langsung, pemfaktoran atau mengalikan dengan faktor sekawan.
Substitusi Langsung
Teknik terbaik memulai perhitungan \(\mathrm{_{x \to c}^{lim}}\) f(x) yakni dengan mensubstitusikan x = c ke fungsi f(x) atau dengan kata lain memilih nilai f(c). Selama f(c) terdefinisi atau ada nilainya (bukan ialah bentuk dukungan dengan nol), maka f(c) yakni nilai limit yang kita cari.misal 1
Hitunglah \(\mathrm{_{x \to 2}^{lim}}\) (x3 - 4x + 1)
Jawab :
\(\mathrm{_{x \to 2}^{lim}}\) (x3 - 4x + 1) = 23 - 4.2 + 1 = 1
misal 2
Hitunglah \(\mathrm{_{x \to 3}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\)
Jawab :
\(\mathrm{_{x \to 3}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\) = \(\frac{3^{2}\,+\,3\,-\,6}{3\,-\,2}\) = 6
misal 3
Hitunglah \(\mathrm{_{x \to 1}^{lim}\,\frac{\sqrt{x^{2}+\,8}\,-\,3}{x\,+\,1}}\)
Jawab :
\(\mathrm{_{x \to 1}^{lim}\,\frac{\sqrt{x^{2}+\,8}\,-\,3}{x\,+\,1}}\) = \(\mathrm{\frac{\sqrt{1^{2}+\,8}\,-\,3}{1\,+\,1}}\) = \(\frac{0}{2}\) = 0
Pemfaktoran / Mengalikan Faktor Sekawan
Jika dengan subsitusi eksklusif diperoleh bentuk \(\frac{0}{0}\), maka perhitungan limit kita alihkan dengan cara pemfaktoran ataupun mengalikan dengan faktor sekawan.Misalkan f(x) = \(\mathrm{\frac{(x\,-\,1)(x\,+\,2)}{x\,-\,1}}\)
Untuk x = 1 maka x - 1 = 0, akhirnya f(x) = 0/0.
Untuk x ≠ 1 maka x - 1 ≠ 0, akhirnya f(x) = x + 2, sanggup kita tulis $$\mathrm{\frac{(x-1)(x+2)}{x-1}=x+2,\;\;ketika\; x}\neq 1$$ Karena limit spesialuntuk mengamati nilai fungsi f ketika x mendekati c (x ≠ c), akhirnya $$\mathrm{\lim_{x\rightarrow 1}\frac{(x-1)(x+2)}{x-1}=\lim_{x\rightarrow 1} (x+2)}$$Secara tidak langsung, uraian diatas mengambarkan kepada kita bahwa ketika bekerja dengan limit, kita diizinkan mengeliminasi atau mencoret faktor yang sama pada pembilang dan penyebut tanpa harus khawatir melanggar hukum dengan mencoret nol.
misal 4
Hitunglah \(\mathrm{_{x \to 2}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{x \to 2}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\) = \(\mathrm{_{x \to 2}^{lim}\,\frac{(x\,{\color{red}\not}-\,2)(x\,+\,3)}{x\,{\color{red}\not}-\,2}}\)
\(\mathrm{_{x \to 2}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\) = \(\mathrm{_{x \to 2}^{lim}}\) (x + 3)
\(\mathrm{_{x \to 2}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\) = 2 + 3
\(\mathrm{_{x \to 2}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\) = 5
misal 5
Hitunglah \(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{2 {\color{red}\not}x}{{\color{red}\not}x(4\,-\,x)}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{2 }{4\,-\,x}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\) = \(\frac{2}{4\,-\,0}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\) = \(\frac{1}{2}\)
misal 6
Hitunglah \(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{(2x\,-\,1)(x\,{\color{red}\not}-\,1)}{(x\,+\,4)(x\,{\color{red}\not}-\,1)}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{2x\,-\,1}{x\,+\,4}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\) = \(\mathrm{\frac{2.1\,-\,1}{1\,+\,4}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\) = \(\mathrm{\frac{1}{5}}\)
misal 7
Hitunglah \(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\) = \(\mathrm{_{h \to 0}^{lim}\,\frac{4\,+\,4h\,+\,h^{2}-4}{h}}\)
\(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\) = \(\mathrm{_{h \to 0}^{lim}\,\frac{{\color{red}\not}h(4\,+\,h)}{{\color{red}\not}h}}\)
\(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\) = \(\mathrm{_{h \to 0}^{lim}}\) (4 + h)
\(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\) = 4 + 0
\(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\) = 4
misal 8
Diketahui f(x) = ax + b dengan a dan b konstan. Hitunglah \(\mathrm{_{h \to 0}^{lim}\,\frac{f(x\,+\,h)\,-\,f(x)}{h}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
f(x) = ax + b
f(x + h) = a(x + h) + b
\(\mathrm{_{h \to 0}^{lim}\,\frac{f(x\,+\,h)\,-\,f(x)}{h}}\) = \(\mathrm{_{h \to 0}^{lim}\,\frac{a(x\,+\,h)\,+\,b\,-\,(ax\,+\,b)}{h}}\)
\(\mathrm{_{h \to 0}^{lim}\,\frac{f(x\,+\,h)\,-\,f(x)}{h}}\) = \(\mathrm{_{h \to 0}^{lim}\,\frac{ax\,+\,ah\,+\,b\,-\,ax\,-\,b}{h}}\)
\(\mathrm{_{h \to 0}^{lim}\,\frac{f(x\,+\,h)\,-\,f(x)}{h}}\) = \(\mathrm{_{h \to 0}^{lim}\,\frac{a{\color{red}\not}h}{{\color{red}\not}h}}\)
\(\mathrm{_{h \to 0}^{lim}\,\frac{f(x\,+\,h)\,-\,f(x)}{h}}\) = \(\mathrm{_{h \to 0}^{lim}}\) a
\(\mathrm{_{h \to 0}^{lim}\,\frac{f(x\,+\,h)\,-\,f(x)}{h}}\) = a
misal 9
Hitunglah \(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
Faktorkan t³ - 8 dengan memakai sifat
a³ - b³ = (a - b)(a² + ab + b²)
t³ - 8 = t³ - 2³ = (t - 2)(t² + 2t + 4)
\(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\) = \(\mathrm{_{t \to 2}^{lim}\,\frac{t\,{\color{red}\not}-\,2}{(t\,{\color{red}\not}-\,2)(t^{2}+\,2t\,+\,4)}}\)
\(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\) = \(\mathrm{_{t \to 2}^{lim}\,\frac{1}{t^{2}+\,2t\,+\,4}}\)
\(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\) = \(\mathrm{\frac{1}{(2)^{2}+\,2(2)\,+\,4}}\)
\(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\) = \(\frac{1}{16}\)
misal 10
Hitunglah \(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
Faktorkan x⁴ - 1 dengan memakai sifat
a² - b² = (a - b)(a + b)
x⁴ - 1 = (x²)2 - (1)2 = (x² - 1)(x² + 1)
x⁴ - 1 = (x²)² - (1)² = (x + 1)(x - 1)(x² + 1)
Faktorkan x³ + 1 dengan memakai sifat
a³ + b³ = (a + b)(a² - ab + b²)
x³ + 1 = x³ + 1³ = (x + 1)(x² - x + 1)
\(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\) = \(\mathrm{_{x \to -1}^{lim}\,\frac{(x\,{\color{red}\not}+\,1)(x\,-\,1)(x^{2}+\,1)}{(x\,{\color{red}\not}+\,1)(x^{2}-\,x\,+\,1)}}\)
\(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\) = \(\mathrm{_{x \to -1}^{lim}\,\frac{(x\,-\,1)(x^{2}+\,1)}{x^{2}-\,x\,+\,1}}\)
\(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\) = \(\frac{(-1\,-\,1)((-1)^{2}+\,1)}{(-1)^{2}-\,(-1)\,+\,1}\)
\(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\) = \(-\frac{4}{3}\)
misal 11
Hitunglah \(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
Faktorkan x - 9 dengan memakai sifat
a - b = (√a - √b)(√a + √b)
x - 9 = (√x - 3)(√x + 3)
\(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\) = \(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}{\color{red}\not}-\,3}{\left ( \sqrt{x}{\color{red}\not}-\,3 \right )\left ( \sqrt{x}+\,3 \right )}}\)
\(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\) = \(\mathrm{_{x \to 9}^{lim}\,\frac{1}{\sqrt{x}+\,3}}\)
\(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\) = \(\mathrm{\frac{1}{\sqrt{9}+\,3}}\)
\(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\) = \(\frac{1}{6}\)
misal 12
Hitunglah \(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
Faktorkan x - 8 dengan memakai sifat
a - b = (∛a - ∛b)(∛a² + ∛ab + ∛b²)
x - 8 = (∛x - 2)(∛x² + ∛8x + 4)
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = \(\mathrm{_{x \to 8}^{lim}\,\frac{\left (\sqrt[3]{\mathrm{x}}{\color{red}\not}\,-\,2 \right )\left ( \sqrt[3]{\mathrm{x}^{2}}\,+\,\sqrt[3]{8\mathrm{x}}\,+\,4 \right )}{\sqrt[3]{\mathrm{x}}{\color{red}\not}\,-\,2}}\)
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = \(\mathrm{_{x \to 8}^{lim}}\) (∛x² + ∛(8x) + 4)
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = ∛8² + ∛(8.8) + 4
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = 4 + 4 + 4
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = 12
Jika fungsi susah untuk difaktorkan (biasanya fungsi-fungsi yang memuat tanda akar), kita sanggup mencoba alternatif lain, yaitu dengan mengalikan faktor sekawan.
misal 13
Hitunglah \(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}\cdot \frac{\sqrt{t^{2}+\,7}\,+\,4}{\sqrt{t^{2}+\,7}\,+\,4}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{t^{2}+\,7\,-\,16}{\left (t\,-\,3 \right )\left ( \sqrt{t^{2}+\,7}\,+\,4 \right )}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{(t\,{\color{red}\not}-\,3)(t\,+\,3)}{\left (t\,{\color{red}\not}-\,3 \right )\left ( \sqrt{t^{2}+\,7}\,+\,4 \right )}}\)
misal 12
Hitunglah \(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
Faktorkan x - 8 dengan memakai sifat
a - b = (∛a - ∛b)(∛a² + ∛ab + ∛b²)
x - 8 = (∛x - 2)(∛x² + ∛8x + 4)
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = \(\mathrm{_{x \to 8}^{lim}\,\frac{\left (\sqrt[3]{\mathrm{x}}{\color{red}\not}\,-\,2 \right )\left ( \sqrt[3]{\mathrm{x}^{2}}\,+\,\sqrt[3]{8\mathrm{x}}\,+\,4 \right )}{\sqrt[3]{\mathrm{x}}{\color{red}\not}\,-\,2}}\)
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = \(\mathrm{_{x \to 8}^{lim}}\) (∛x² + ∛(8x) + 4)
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = ∛8² + ∛(8.8) + 4
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = 4 + 4 + 4
\(\mathrm{_{x \to 8}^{lim}\,\frac{x\,-\,8}{\sqrt[3]{\mathrm{x}}-\,2}}\) = 12
Jika fungsi susah untuk difaktorkan (biasanya fungsi-fungsi yang memuat tanda akar), kita sanggup mencoba alternatif lain, yaitu dengan mengalikan faktor sekawan.
misal 13
Hitunglah \(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}\cdot \frac{\sqrt{t^{2}+\,7}\,+\,4}{\sqrt{t^{2}+\,7}\,+\,4}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{t^{2}+\,7\,-\,16}{\left (t\,-\,3 \right )\left ( \sqrt{t^{2}+\,7}\,+\,4 \right )}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{(t\,{\color{red}\not}-\,3)(t\,+\,3)}{\left (t\,{\color{red}\not}-\,3 \right )\left ( \sqrt{t^{2}+\,7}\,+\,4 \right )}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{t\,+\,3}{\sqrt{t^{2}+\,7}\,+\,4}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{\frac{3\,+\,3}{\sqrt{3^{2}+\,7}\,+\,4}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\frac{3}{4}\)
misal 14
Hitunglah \(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
misal 14
Hitunglah \(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}\cdot \frac{{\color{Red} 2\,+\sqrt{5\,-\,x^{2}}}}{{\color{Red} 2\,+\sqrt{5\,-\,x^{2}}}}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{\left (1\,-\,x^{2} \right ) \left ( 2\,+\sqrt{5\,-\,x^{2}} \right )}{4\,-(5\,-\,x^{2})}}\)\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{\left (1\,{\color{red}\not}-\,x^{2} \right ) \left ( 2\,+\sqrt{5\,-\,x^{2}} \right )}{-(1\,{\color{red}\not}-\,x^{2})}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{-\,_{x \to 1}^{lim}\,\left (2\,+\sqrt{5\,-\,x^{2}} \right )}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{-\left (2\,+\sqrt{5\,-\,1^{2}} \right )}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = -4
misal 15
Hitunglah \(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}\cdot \frac{ \sqrt{4x\,+\,2}\,+\,\sqrt{2}}{ \sqrt{4x\,+\,2}\,+\,\sqrt{2}}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{(4x\,+\,2)\,-\,2}{x\left (\sqrt{4x\,+\,2}\,+\sqrt{2} \right )}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{4{\color{red}\not}x}{{\color{red}\not}x\left (\sqrt{4x\,+\,2}\,+\sqrt{2} \right )}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{4}{\sqrt{4x\,+\,2}\,+\sqrt{2}}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{\frac{4}{\sqrt{4.0\,+\,2}\,+\sqrt{2}}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{\frac{4}{2\sqrt{2}}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = √2
misal 16
Hitunglah \(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\)
Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{_{u \to 1}^{lim}\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}\cdot \frac{ \sqrt{2u\,+\,1}\,+\,\sqrt{3u}}{ \sqrt{2u\,+\,1}\,+\,\sqrt{3u}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{_{u \to 1}^{lim}\,\frac{(2u\,+\,1)\,-\,(3u)}{(u\,-\,1){ \left (\sqrt{2u\,+\,1}\,+\,\sqrt{3u} \right )}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{_{u \to 1}^{lim}\,\frac{-(u\,{\color{red}\not}-\,1)}{(u\,{\color{red}\not}-\,1){ \left (\sqrt{2u\,+\,1}\,+\,\sqrt{3u} \right )}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{_{u \to 1}^{lim}\,\frac{-1}{{\sqrt{2u\,+\,1}\,+\,\sqrt{3u}}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{\frac{-1}{{\sqrt{2.1\,+\,1}\,+\,\sqrt{3.1}}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{-\frac{1}{{ 2\sqrt{3}}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{-\frac{1}{{ 6}}}\)√3
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