Limit Fungsi Aljabar


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

Substitusi Langsung

Cara terbaik memulai perhitungan  \(\mathrm{_{x \to c}^{lim}}\) f(x) adalah dengan mensubstitusikan x = c ke fungsi f(x) atau dengan kata lain menentukan nilai f(c). Selama f(c)  terdefinisi atau ada nilainya (bukan merupakan bentuk pembagian dengan nol), maka f(c) adalah 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 langsung 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, akibatnya f(x) = 0/0.
Untuk x ≠ 1 maka x - 1 ≠ 0, akibatnya f(x) = x + 2, dapat kita tulis  $$\mathrm{\frac{(x-1)(x+2)}{x-1}=x+2,\;\;ketika\; x}\neq 1$$Karena limit hanya mengamati nilai fungsi f saat x mendekati c (x ≠ c), akibatnya $$\mathrm{\lim_{x\rightarrow 1}\frac{(x-1)(x+2)}{x-1}=\lim_{x\rightarrow 1} (x+2)}$$Secara tidak langsung, uraian diatas menjelaskan kepada kita bahwa saat bekerja dengan limit, kita diizinkan mengeliminasi atau mencoret faktor yang sama pada pembilang dan penyebut tanpa harus khawatir melanggar aturan dengan mencoret nol.


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

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

\(\mathrm{_{x \to 2}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\) = \(\mathrm{_{x \to 2}^{lim}\,\frac{(x\,{\color{red}\not}-\,2)(x\,+\,3)}{x\,{\color{red}\not}-\,2}}\)
\(\mathrm{_{x \to 2}^{lim}\,\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}\) = \(\mathrm{_{x \to 2}^{lim}}\) (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 langsung ⇒ \(\frac{0}{0}\)

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


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

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

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


 Contoh 7 
Hitunglah  \(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\)

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

\(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\) = \(\mathrm{_{h \to 0}^{lim}\,\frac{4\,+\,4h\,+\,h^{2}-4}{h}}\)
\(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\) = \(\mathrm{_{h \to 0}^{lim}\,\frac{{\color{red}\not}h(4\,+\,h)}{{\color{red}\not}h}}\)
\(\mathrm{_{h \to 0}^{lim}\,\frac{(2\,+\,h)^{2}-\,4}{h}}\) = \(\mathrm{_{h \to 0}^{lim}}\) (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 langsung ⇒ \(\frac{0}{0}\)

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 langsung ⇒ \(\frac{0}{0}\)

Faktorkan t³ - 8 dengan menggunakan 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 langsung ⇒ \(\frac{0}{0}\)

Faktorkan x⁴ - 1 dengan menggunakan 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 menggunakan 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 langsung ⇒ \(\frac{0}{0}\)

Faktorkan x - 9 dengan menggunakan 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 langsung ⇒ \(\frac{0}{0}\)

Faktorkan x - 8 dengan menggunakan 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 dapat 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 langsung ⇒ \(\frac{0}{0}\)

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


 Contoh 14 
Hitunglah  \(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\)

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

\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}\cdot \frac{{\color{Red} 2\,+\sqrt{5\,-\,x^{2}}}}{{\color{Red} 2\,+\sqrt{5\,-\,x^{2}}}}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{\left (1\,-\,x^{2}  \right ) \left (  2\,+\sqrt{5\,-\,x^{2}} \right )}{4\,-(5\,-\,x^{2})}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{_{x \to 1}^{lim}\,\frac{\left (1\,{\color{red}\not}-\,x^{2}  \right ) \left (  2\,+\sqrt{5\,-\,x^{2}} \right )}{-(1\,{\color{red}\not}-\,x^{2})}}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{-\,_{x \to 1}^{lim}\,\left (2\,+\sqrt{5\,-\,x^{2}}  \right )}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = \(\mathrm{-\left (2\,+\sqrt{5\,-\,1^{2}}  \right )}\)
\(\mathrm{_{x \to 1}^{lim}\,\frac{1\,-\,x^{2}}{2\,-\sqrt{5\,-\,x^{2}}}}\) = -4


 Contoh 15 
Hitunglah  \(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\)

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

\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}\cdot \frac{ \sqrt{4x\,+\,2}\,+\,\sqrt{2}}{ \sqrt{4x\,+\,2}\,+\,\sqrt{2}}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{(4x\,+\,2)\,-\,2}{x\left (\sqrt{4x\,+\,2}\,+\sqrt{2}  \right )}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{4{\color{red}\not}x}{{\color{red}\not}x\left (\sqrt{4x\,+\,2}\,+\sqrt{2}  \right )}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{_{x \to 0}^{lim}\,\frac{4}{\sqrt{4x\,+\,2}\,+\sqrt{2}}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{\frac{4}{\sqrt{4.0\,+\,2}\,+\sqrt{2}}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = \(\mathrm{\frac{4}{2\sqrt{2}}}\)
\(\mathrm{_{x \to 0}^{lim}\,\frac{\sqrt{4x\,+\,2}\,-\,\sqrt{2}}{x}}\) = √2


 Contoh 16 
Hitunglah  \(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\)

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

\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{_{u \to 1}^{lim}\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}\cdot \frac{ \sqrt{2u\,+\,1}\,+\,\sqrt{3u}}{ \sqrt{2u\,+\,1}\,+\,\sqrt{3u}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{_{u \to 1}^{lim}\,\frac{(2u\,+\,1)\,-\,(3u)}{(u\,-\,1){ \left (\sqrt{2u\,+\,1}\,+\,\sqrt{3u}  \right )}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{_{u \to 1}^{lim}\,\frac{-(u\,{\color{red}\not}-\,1)}{(u\,{\color{red}\not}-\,1){ \left (\sqrt{2u\,+\,1}\,+\,\sqrt{3u}  \right )}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{_{u \to 1}^{lim}\,\frac{-1}{{\sqrt{2u\,+\,1}\,+\,\sqrt{3u}}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{\frac{-1}{{\sqrt{2.1\,+\,1}\,+\,\sqrt{3.1}}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{-\frac{1}{{ 2\sqrt{3}}}}\)
\(\mathrm{_{u \to 1}^{lim}\,\frac{\sqrt{2u\,+\,1}\,-\,\sqrt{3u}}{u\,-\,1}}\) = \(\mathrm{-\frac{1}{{ 6}}}\)√3


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