Limit Fungsi Aljabar

Pada artikel ini kita akan mempelajari cara menghitung nilai limit dari fungsi-fungsi aljabar, khususnya polinom dan rasional. Adapun metode yang dipakai yaitu substitusi langsung, pemfaktoran atau mengalikan dengan faktor sekawan.

Substitusi Langsung

Cara terbaik memulai perhitungan  \(\mathrm{_{x \to c}^{lim}}\) f(x) yaitu dengan mensubstitusikan x = c ke fungsi f(x) atau dengan kata lain memilih nilai f(c). Selama f(c)  terdefinisi atau ada nilainya (bukan merupakan bentuk pembagian dengan nol), maka f(c) yaitu nilai limit yang kita cari.

 Contoh 1 
Hitunglah  \(\mathrm{_{x \to 2}^{lim}}\) (x³ - 4x + 1)

Jawab :
\(\mathrm{_{x \to 2}^{lim}}\) (x³ - 4x + 1) = 2³ - 4.2 + 1 = 1


 Contoh 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


 Contoh 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, risikonya f(x) = 0/0.
Untuk x ≠ 1 maka x - 1 ≠ 0, risikonya f(x) = x + 2, sanggup kita tulis  $$\mathrm{\frac{(x-1)(x+2)}{x-1}=x+2,\;\;ketika\; x}\neq 1$$ Karena limit hanya mengamati nilai fungsi f ketika x mendekati c (x ≠ c), risikonya $$\mathrm{\lim_{x\rightarrow 1}\frac{(x-1)(x+2)}{x-1}=\lim_{x\rightarrow 1} (x+2)}$$Secara tidak langsung, uraian diatas menjelaskan 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.


 Contoh 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


 Contoh 5 
Hitunglah  \(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\)

Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)

\(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{2 {\color{red}\not}x}{{\color{red}\not}x(4\,-\,x)}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{2 }{4\,-\,x}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\) = \(\frac{2}{4\,-\,0}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{2x}{4x\,-\,x^{2}}}\) = \(\frac{1}{2}\)


 Contoh 6 
Hitunglah  \(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\)

Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)

\(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{(2x\,-\,1)(x\,{\color{red}\not}-\,1)}{(x\,+\,4)(x\,{\color{red}\not}-\,1)}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{2x\,-\,1}{x\,+\,4}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\) = \(\mathrm{\frac{2.1\,-\,1}{1\,+\,4}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{2x^{2}-\,3x\,+\,1}{x^{2}+\,3x\,-\,4}}\) = \(\mathrm{\frac{1}{5}}\)


 Contoh 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


 Contoh 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


 Contoh 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)

\(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\) = \(\mathrm{_{t \to 2}^{lim}\,\frac{t\,{\color{red}\not}-\,2}{(t\,{\color{red}\not}-\,2)(t^{2}+\,2t\,+\,4)}}\)
\(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\) = \(\mathrm{_{t \to 2}^{lim}\,\frac{1}{t^{2}+\,2t\,+\,4}}\)
\(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\) = \(\mathrm{\frac{1}{(2)^{2}+\,2(2)\,+\,4}}\)
\(\mathrm{_{t \to 2}^{lim}\,\frac{t\,-\,2}{t^{3}-\,8}}\) = \(\frac{1}{16}\)


 Contoh 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)

\(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\) = \(\mathrm{_{x \to -1}^{lim}\,\frac{(x\,{\color{red}\not}+\,1)(x\,-\,1)(x^{2}+\,1)}{(x\,{\color{red}\not}+\,1)(x^{2}-\,x\,+\,1)}}\)
\(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\) = \(\mathrm{_{x \to -1}^{lim}\,\frac{(x\,-\,1)(x^{2}+\,1)}{x^{2}-\,x\,+\,1}}\)
\(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\) = \(\frac{(-1\,-\,1)((-1)^{2}+\,1)}{(-1)^{2}-\,(-1)\,+\,1}\)
\(\mathrm{_{x \to -1}^{lim}\,\frac{x^{4}-\,1}{x^{3}+\,1}}\) = \(-\frac{4}{3}\)


 Contoh 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)

\(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\) = \(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}{\color{red}\not}-\,3}{\left ( \sqrt{x}{\color{red}\not}-\,3 \right )\left ( \sqrt{x}+\,3 \right )}}\)
\(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\) = \(\mathrm{_{x \to 9}^{lim}\,\frac{1}{\sqrt{x}+\,3}}\)
\(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\) = \(\mathrm{\frac{1}{\sqrt{9}+\,3}}\)

\(\mathrm{_{x \to 9}^{lim}\,\frac{\sqrt{x}-\,3}{x\,-\,9}}\) = \(\frac{1}{6}\)


 Contoh 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 sulit untuk difaktorkan (biasanya fungsi-fungsi yang memuat tanda akar), kita sanggup mencoba alternatif lain, yaitu dengan mengalikan faktor sekawan.

 Contoh 13 
Hitunglah  \(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\)

Jawab :
Substitusi eksklusif ⇒ \(\frac{0}{0}\)

\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}\cdot \frac{\sqrt{t^{2}+\,7}\,+\,4}{\sqrt{t^{2}+\,7}\,+\,4}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{t^{2}+\,7\,-\,16}{\left (t\,-\,3  \right )\left ( \sqrt{t^{2}+\,7}\,+\,4 \right )}}\)
\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{(t\,{\color{red}\not}-\,3)(t\,+\,3)}{\left (t\,{\color{red}\not}-\,3  \right )\left ( \sqrt{t^{2}+\,7}\,+\,4 \right )}}\)

\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{_{t \to 3}^{lim}\,\frac{t\,+\,3}{\sqrt{t^{2}+\,7}\,+\,4}}\)

\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\mathrm{\frac{3\,+\,3}{\sqrt{3^{2}+\,7}\,+\,4}}\)

\(\mathrm{_{t \to 3}^{lim}\,\frac{\sqrt{t^{2}+\,7}\,-\,4}{t\,-\,3}}\) = \(\frac{3}{4}\)


 Contoh 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


 Contoh 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


 Contoh 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



Sumber http://smatika.blogspot.com

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