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Bahas Soal Matematika » Limit dan Kekontinuan › Contoh Soal Limit Fungsi Aljabar Metode Perkalian Akar Sekawan
Joki Tugas Matematika Murah, Hanya Rp10-50 Ribu. Hub. WA: 0812-5632-4552
Contoh Soal Limit Fungsi Aljabar Metode Perkalian Akar Sekawan
Sering kali dalam soal limit fungsi aljabar terdapat bentuk akar yang terkadang menyulitkan proses pencarian nilai limitnya. Jika itu terjadi, kita bisa coba selesaikan soal limit tersebut dengan metode perkalian dengan akar sekawan atau metode mengalikan dengan faktor sekawan.
Ini dari metode ini yaitu mengalikan fungsi pada limit dengan akar sekawannya. Tentu saja metode ini kita gunakan apabila metode substitusi menghasilkan nilai limit yang tak terdefinisi atau merupakan bentuk tak tentu \(0/0\) atau \(∞/∞\).
Untuk lebih jelasnya, berikut ini kita telah siapkan sejumlah contoh soal limit fungsi aljabar yang penyelesaiannya bisa menggunakan metode perkalian dengan akar sekawan.
Contoh 1: EBTANAS SMA IPA 1999
Nilai \( \displaystyle \lim_{x \to 2} \ \frac{x-2}{\sqrt{x+7}-3} = \cdots \)
- -2
- -2/3
- 0
- 6
- 12
Perhatikan bahwa fungsi penyebut pada limit mempunyai bentuk akar sehingga kita bisa coba selesaikan limit yang diberikan pada soal menggunakan metode perkalian akar sekawan. Akar sekawannya yaitu \( \sqrt{x+7}+3 \). Berikut hasil yang diperoleh:
\begin{aligned} \lim_{x \to 2} \ \frac{x-2}{\sqrt{x+7}-3} &= \lim_{x \to 2} \ \frac{x-2}{\sqrt{x+7}-3} \times \frac{\sqrt{x+7}+3}{\sqrt{x+7}+3} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(\sqrt{x+7}+3)}{(x+7)-9} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(\sqrt{x+7}+3)}{(x-2)} \\[8pt] &= \lim_{x \to 2} \ (\sqrt{x+7}+3) \\[8pt] &= \sqrt{2+7}+3 = \sqrt{9} + 3 = 6 \end{aligned}
Jawaban D.
Contoh 2: SPMB 2007
Nilai \( \displaystyle \lim_{x \to 1} \ \frac{1-\sqrt{3x-2}}{1-x} = \cdots \)
- \( -1 \frac{1}{2} \)
- -1
- 0
- 1
- \( 1 \frac{1}{2} \)
Sama seperti pada Contoh 1 di atas, terdapat bentuk akar pada fungsi yang diberikan dalam limit, hanya saja dalam soal ini bentuk akarnya ada pada pembilang. Oleh karena itu, kita bisa coba selesaikan limit ini dengan mengalikan bentuk sekawannya yaitu \( (1+\sqrt{3x-2}) \). Berikut hasil yang diperoleh:
\begin{aligned} \lim_{x \to 1} \ \frac{1-\sqrt{3x-2}}{1-x} &= \lim_{x \to 1} \ \frac{1-\sqrt{3x-2}}{1-x} \times \frac{1+\sqrt{3x-2}}{1+\sqrt{3x-2}} \\[8pt] &= \lim_{x \to 1} \ \frac{1-(3x-2)}{(1-x)(1+\sqrt{3x-2})} \\[8pt] &= \lim_{x \to 1} \ \frac{3(1-x)}{(1-x)(1+\sqrt{3x-2})} \\[8pt] &= \lim_{x \to 1} \ \frac{3}{(1+\sqrt{3x-2})} \\[8pt] &= \frac{3}{1+\sqrt{3(1)-2}} = \frac{3}{2} = 1 \frac{1}{2} \end{aligned}
Jawaban E.
Contoh 3: SPMB 2007
Nilai \( \displaystyle \lim_{x \to 1} \ \frac{x-1}{\sqrt{x+3}-2} = \cdots \)
- 1/4
- 1/2
- 1
- 2
- 4
\begin{aligned} \lim_{x \to 1} \ \frac{x-1}{\sqrt{x+3}-2} &= \lim_{x \to 1} \ \frac{x-1}{\sqrt{x+3}-2} \times \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2} \\[8pt] &= \lim_{x \to 1} \ \frac{(x-1)(\sqrt{x+3}+2)}{(x+3)-4} \\[8pt] &= \lim_{x \to 1} \ \frac{(x-1)(\sqrt{x+3}+2)}{x-1} \\[8pt] &= \lim_{x \to 1} \ (\sqrt{x+3}+2) \\[8pt] &= \sqrt{1+3}+2 = 4 \end{aligned}
Jawaban E.
Contoh 4: SNMPTN 2008
Nilai \( \displaystyle \lim_{x \to 2} \ \frac{\sqrt{3x-2}-2}{2x-4} = \cdots \)
- 0
- 3/8
- 3/4
- 1
- \( 1 \frac{1}{2} \)
\begin{aligned} \lim_{x \to 2} \ \frac{\sqrt{3x-2}-2}{2x-4} &= \lim_{x \to 2} \ \frac{\sqrt{3x-2}-2}{2x-4} \times \frac{\sqrt{3x-2}+2}{\sqrt{3x-2}+2} \\[8pt] &= \lim_{x \to 2} \ \frac{(3x-2)-4}{(2x-4)(\sqrt{3x-2}+2)} \\[8pt] &= \lim_{x \to 2} \ \frac{3(x-2)}{2(x-2)(\sqrt{3x-2}+2)} \\[8pt] &= \lim_{x \to 2} \ \frac{3}{2(\sqrt{3x-2}+2)} \\[8pt] &= \frac{3}{2(\sqrt{3(2)-2}+2)} \\[8pt] &= \frac{3}{8} \end{aligned}
Jawaban B.
Contoh 5: UM UGM 2007
Nilai \( \displaystyle \lim_{x \to 2} \ \frac{\sqrt{x^2+5}-3}{x^2-2x} = \cdots \)
- 0
- 1/3
- 1/2
- 3/4
- ∞
\begin{aligned} \lim_{x \to 2} \ \frac{\sqrt{x^2+5}-3}{x^2-2x} &= \lim_{x \to 2} \ \frac{\sqrt{x^2+5}-3}{x^2-2x} \times \frac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3} \\[8pt] &= \lim_{x \to 2} \ \frac{(x^2+5)-9}{(x^2-2x)(\sqrt{x^2+5}+3)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x^2-4)}{(x^2-2x)(\sqrt{x^2+5}+3)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x+2)}{x(x-2)(\sqrt{x^2+5}+3)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x+2)}{x \ (\sqrt{x^2+5}+3)} = \frac{(2+2)}{2 \ (\sqrt{2^2+5}+3)} \\[8pt] &= \frac{4}{2(\sqrt{9}+3)} = \frac{4}{12} = \frac{1}{3} \end{aligned}
Jawaban B.
Contoh 6: UMB PTN 2012
Nilai \( \displaystyle \lim_{x \to 3} \ \frac{x-3}{3-\sqrt{x+6}} = \cdots \)
- -6
- -3
- 0
- 3
- 6
\begin{aligned} \lim_{x \to 3} \ \frac{x-3}{3-\sqrt{x+6}} &= \lim_{x \to 3} \ \frac{x-3}{3-\sqrt{x+6}} \times \frac{3+\sqrt{x+6}}{3+\sqrt{x+6}} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(3+\sqrt{x+6})}{9-(x+6)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(3+\sqrt{x+6})}{9-(x+6)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(3+\sqrt{x+6})}{-(x-3)} \\[8pt] &= \lim_{x \to 3} \ \frac{(3+\sqrt{x+6})}{-1} \\[8pt] &= \frac{3+\sqrt{3+6}}{-1} = -6 \end{aligned}
Jawaban A.
Contoh 7: SPMB 2006
Nilai \( \displaystyle \lim_{x \to 7} \ \frac{\sqrt{x}(x-7)}{\sqrt{x}-\sqrt{7}} = \cdots \)
- 14
- 7
- \( 2 \sqrt{7} \)
- \( \sqrt{7} \)
- \( \frac{1}{2} \sqrt{7} \)
\begin{aligned} \lim_{x \to 7} \ \frac{\sqrt{x}(x-7)}{\sqrt{x}-\sqrt{7}} &= \lim_{x \to 7} \ \frac{\sqrt{x}(x-7)}{\sqrt{x}-\sqrt{7}} \times \frac{\sqrt{x}+\sqrt{7}}{\sqrt{x}+\sqrt{7}} \\[8pt] &= \lim_{x \to 7} \ \frac{\sqrt{x}(x-7)(\sqrt{x}+\sqrt{7})}{x-7} \\[8pt] &= \lim_{x \to 7} \ \sqrt{x} \ (\sqrt{x}+\sqrt{7}) \\[8pt] &= \sqrt{7} \ (\sqrt{7}+\sqrt{7}) \\[8pt] &= \sqrt{7} \ (2\sqrt{7}) = 14 \end{aligned}
Jawaban A.
Contoh 8: SPMB 2007
Nilai \( \displaystyle \lim_{x \to 4} \ \frac{x^2-16}{5-\sqrt{x^2+9}} = \cdots \)
- -20
- -10
- 0
- 8
- 20
\begin{aligned} \lim_{x \to 4} \ \frac{x^2-16}{5-\sqrt{x^2+9}} &= \lim_{x \to 4} \ \frac{x^2-16}{5-\sqrt{x^2+9}} \times \frac{5+\sqrt{x^2+9}}{5+\sqrt{x^2+9}} \\[8pt] &= \lim_{x \to 4} \ \frac{(x^2-16)(5+\sqrt{x^2+9})}{25-(x^2+9)} \\[8pt] &= \lim_{x \to 4} \ \frac{(x^2-16)(5+\sqrt{x^2+9})}{-(x^2-16)} \\[8pt] &= \lim_{x \to 4} \ -(5+\sqrt{x^2+9}) \\[8pt] &= -(5+\sqrt{4^2+9}) \\[8pt] &= -10 \end{aligned}
Jawaban B.
Contoh 9: SPMB 2007
Nilai \( \displaystyle \lim_{x \to 5} \ \frac{x^2-25}{\sqrt{x^2+24}-7} = \cdots \)
- 0
- 5
- 7
- 14
- 18
\begin{aligned} \lim_{x \to 5} \ \frac{x^2-25}{\sqrt{x^2+24}-7} &= \lim_{x \to 5} \ \frac{x^2-25}{\sqrt{x^2+24}-7} \times \frac{\sqrt{x^2+24}+7}{\sqrt{x^2+24}+7} \\[8pt] &= \lim_{x \to 5} \ \frac{(x^2-25)(\sqrt{x^2+24}+7)}{(x^2+24)-49} \\[8pt] &= \lim_{x \to 5} \ \frac{(x^2-25)(\sqrt{x^2+24}+7)}{x^2-25} \\[8pt] &= \lim_{x \to 5} \ (\sqrt{x^2+24}+7) \\[8pt] &= \sqrt{5^2 + 24} + 7 = \sqrt{49} + 7 \\[8pt] &= 14 \end{aligned}
Jawaban D.
Contoh 10: UM UGM 2006
Nilai \( \displaystyle f(x)= \frac{1-x}{2-\sqrt{x^2+3}} \), maka \( \displaystyle \lim_{x \to 1} \ f(x) = \cdots \)
- 0
- 1/2
- 1
- 2
- 4
\begin{aligned} \lim_{x \to 1} \ \frac{1-x}{2-\sqrt{x^2+3}} &= \lim_{x \to 1} \ \frac{1-x}{2-\sqrt{x^2+3}} \times \frac{2+\sqrt{x^2+3}}{2+\sqrt{x^2+3}} \\[8pt] &= \lim_{x \to 1} \ \frac{(1-x)(2+\sqrt{x^2+3})}{4-(x^2+3)} \\[8pt] &= \lim_{x \to 1} \ \frac{(1-x)(2+\sqrt{x^2+3})}{1-x^2} \\[8pt] &= \lim_{x \to 1} \ \frac{(1-x)(2+\sqrt{x^2+3})}{(1-x)(1+x)} \\[8pt] &= \lim_{x \to 1} \ \frac{(2+\sqrt{x^2+3})}{(1+x)} = \frac{2+\sqrt{1^2+3}}{1+1} \\[8pt] &= \frac{2+2}{2} = 2 \end{aligned}
Jawaban D.
Contoh 11: SPMB 2007
Nilai \( \displaystyle \lim_{x \to 1} \ \frac{(x-1)(\sqrt{x}+1)}{(\sqrt{x}-1)} = \cdots \)
- 0
- 1
- 2
- 4
- 8
\begin{aligned} \lim_{x \to 1} \ \frac{(x-1)(\sqrt{x}+1)}{\sqrt{x}-1} &= \lim_{x \to 1} \ \frac{(x-1)(\sqrt{x}+1)}{\sqrt{x}-1} \times \frac{\sqrt{x}+1}{\sqrt{x}+1} \\[8pt] &= \lim_{x \to 1} \ \frac{(x-1)(\sqrt{x}+1)^2}{x-1} \\[8pt] &= \lim_{x \to 1} \ (\sqrt{x}+1)^2 \\[8pt] &= (\sqrt{1}+1)^2 = 4 \end{aligned}
Jawaban D.
Contoh 12: SPMB 2007
Nilai \( \displaystyle \lim_{x \to 1} \frac{\sqrt{2-x}-x}{x^2-x} = \cdots \)
- \( -1 \frac{1}{2} \)
- -1
- 0
- 1
- \( 1 \frac{1}{2} \)
\begin{aligned} \lim_{x \to 1} \ \frac{\sqrt{2-x}-x}{x^2-x} &= \lim_{x \to 1} \ \frac{\sqrt{2-x}-x}{x^2-x} \times \frac{\sqrt{2-x}+x}{\sqrt{2-x}+x} \\[8pt] &= \lim_{x \to 1} \ \frac{(2-x)-x^2}{(x^2-x)(\sqrt{2-x}+x)} \\[8pt] &= \lim_{x \to 1} \ \frac{-(x^2+x-2)}{(x^2-x)(\sqrt{2-x}+x)} \\[8pt] &= \lim_{x \to 1} \ \frac{-(x+2)(x-1)}{x(x-1)(\sqrt{2-x}+x)} \\[8pt] &= \lim_{x \to 1} \ \frac{-(x+2)}{x \ (\sqrt{2-x}+x)} \\[8pt] &= \frac{-(1+2)}{(1) \ (\sqrt{2-1}+1)} = -\frac{3}{2} = - 1 \frac{1}{2} \end{aligned}
Jawaban A.
Contoh 13: UNBK SMA IPA 2019
Nilai \( \displaystyle \lim_{x \to 2} \ \frac{x^2-6x+8}{3-\sqrt{17-2x^2}} = \cdots \)
- 6
- 3/2
- 0
- -3/2
- -6
\begin{aligned} \lim_{x \to 2} \ \frac{x^2-6x+8}{3-\sqrt{17-2x^2}} &= \lim_{x \to 2} \ \frac{x^2-6x+8}{3-\sqrt{17-2x^2}} \times \frac{3+\sqrt{17-2x^2}}{3+\sqrt{17-2x^2}} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x-4)(3+\sqrt{17-2x^2})}{9-(17-2x^2)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x-4)(3+\sqrt{17-2x^2})}{2x^2-8} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x-4)(3+\sqrt{17-2x^2})}{2(x-2)(x+2)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-4)(3+\sqrt{17-2x^2})}{2(x+2)} \\[8pt] &= \frac{(2-4)(3+\sqrt{17-2(2)^2})}{2(2+2)} \\[8pt] &= \frac{(-2)(3+\sqrt{9})}{8} = -\frac{3}{2} \end{aligned}
Jawaban D.
Contoh 14: UN SMA IPA 2003
Nilai \( \displaystyle \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} = \cdots \)
- 0
- 2
- 3
- 4
- 6
\begin{aligned} \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} &= \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} \times \frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}} \\[8pt] &= \lim_{x \to 2} \ \frac{(4-x^2)(3+\sqrt{x^2+5})}{9-(x^2+5)} \\[8pt] &= \lim_{x \to 2} \ \frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2} \\[8pt] &= \lim_{x \to 2} \ (3+\sqrt{x^2+5}) \\[8pt] &= 3+\sqrt{2^2+5} = 3 + \sqrt{9} \\[8pt] &= 6 \end{aligned}
Jawaban E.
Contoh 15: UN SMA IPA 2019
Nilai \( \displaystyle \lim_{x \to 3} \ \frac{x^2-x-6}{\sqrt{3x^2-2}-5} = \cdots \)
- 0
- 25/9
- 25/6
- 25/3
- ∞
\begin{aligned} \lim_{x \to 3} \ \frac{x^2-x-6}{\sqrt{3x^2-2}-5} &= \lim_{x \to 3} \ \frac{x^2-x-6}{\sqrt{3x^2-2}-5} \times \frac{\sqrt{3x^2-2}+5}{\sqrt{3x^2-2}+5} \\[8pt] &= \lim_{x \to 3} \ \frac{(x^2-x-6)(\sqrt{3x^2-2}+5)}{(3x^2-2)-25} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(x+2)(\sqrt{3x^2-2}+5)}{3(x^2-9)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x-3)(x+2)(\sqrt{3x^2-2}+5)}{3(x-3)(x+3)} \\[8pt] &= \lim_{x \to 3} \ \frac{(x+2)(\sqrt{3x^2-2}+5)}{3(x+3)} \\[8pt] &= \frac{(3+2)(\sqrt{3(3)^2-2}+5)}{3(3+3)} \\[8pt] &= \frac{(5)(\sqrt{25}+5)}{18} = \frac{(5)(10)}{18} = \frac{25}{9} \end{aligned}
Jawaban B.
Contoh 16:
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{x}{2 \csc x \ (1-\sqrt{\cos x})} = \cdots \)
- -2
- -1
- 0
- 1
- 2
\begin{aligned} \lim_{x \to 0} \ \frac{x}{2 \csc x \ (1-\sqrt{\cos x})} &= \lim_{x \to 0} \ \frac{x}{2 \csc x \ (1-\sqrt{\cos x})} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} \\[8pt] &= \lim_{x \to 0} \ \frac{x \ (1+\sqrt{\cos x})}{\frac{2}{\sin x} \ (1-\cos x)} \\[8pt] &= \lim_{x \to 0} \ \frac{x \sin x \ (1+\sqrt{\cos x})}{2 \ (2 \sin^2 \frac{1}{2}x)} \\[8pt] &= \lim_{x \to 0} \ \frac{x}{4 \sin \frac{1}{2}x} \cdot \lim_{x \to 0} \ \frac{\sin x}{\sin \frac{1}{2}x} \cdot \lim_{x \to 0} \ (1+\sqrt{\cos x}) \\[8pt] &= \frac{1}{4 \cdot \frac{1}{2}} \cdot \frac{1}{\frac{1}{2}} \cdot (1+\sqrt{\cos 0}) \\[8pt] &= \frac{1}{2} \cdot 2 \cdot (1+1) = 2 \end{aligned}
Jawaban E.
Contoh 17: EBTANAS SMA IPA 2000
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{x^2}{1-\sqrt{1+x^2}} = \cdots \)
- 2
- 0
- -1
- -2
- -3
\begin{aligned} \lim_{x \to 0} \ \frac{x^2}{1-\sqrt{1+x^2}} &= \lim_{x \to 0} \ \frac{x^2}{1-\sqrt{1+x^2}} \times \frac{1+\sqrt{1+x^2}}{1+\sqrt{1+x^2}} \\[8pt] &= \lim_{x \to 0} \ \frac{x^2(1+\sqrt{1+x^2})}{1-(1+x^2)} \\[8pt] &= \lim_{x \to 0} \ \frac{x^2(1+\sqrt{1+x^2})}{-x^2} \\[8pt] &= \lim_{x \to 0} \ -(1+\sqrt{1+x^2}) \\[8pt] &= -(1+\sqrt{1+0^2}) = -2 \end{aligned}
Jawaban D.
Contoh 18: UMB PTN 2013
Nilai \( \displaystyle \lim_{x \to 2} \ \frac{x^2-2x}{\sqrt{x^2+5}-3} = \cdots \)
- -6
- -3
- 0
- 3
- 6
\begin{aligned} \lim_{x \to 2} \ \frac{x^2-2x}{\sqrt{x^2+5}-3} &= \lim_{x \to 2} \ \frac{x^2-2x}{\sqrt{x^2+5}-3} \times \frac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3} \\[8pt] &= \lim_{x \to 2} \ \frac{(x^2-2x)(\sqrt{x^2+5}+3)}{(x^2+5)-9} \\[8pt] &= \lim_{x \to 2} \ \frac{x(x-2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)} \\[8pt] &= \lim_{x \to 2} \ \frac{x(\sqrt{x^2+5}+3)}{(x+2)} = \frac{2(\sqrt{2^2+5}+3)}{(2+2)} \\[8pt] &= \frac{2(\sqrt{9}+3)}{4} = \frac{12}{4} = 3 \end{aligned}
Jawaban D.
Contoh 19: UM UGM 2013
Jika \( \displaystyle a = \lim_{x \to 2} \ \frac{x^2-4}{2-\sqrt{x+2}}\), maka nilai \((4-a)\) adalah...
- -20
- -12
- -4
- 12
- 20
\begin{aligned} a &= \lim_{x \to 2} \ \frac{x^2-4}{2-\sqrt{x+2}} \\[8pt] &= \lim_{x \to 2} \ \frac{x^2-4}{2-\sqrt{x+2}} \times \frac{2+\sqrt{x+2}}{2+\sqrt{x+2}} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x+2)(2+\sqrt{x+2})}{4-(x+2)} \\[8pt] &= \lim_{x \to 2} \ \frac{(x-2)(x+2)(2+\sqrt{x+2})}{-(x-2)} \\[8pt] &= \lim_{x \to 2} \ -(x+2)(2+\sqrt{x+2}) \\[8pt] &= -(2+2)(2+\sqrt{2+2}) \\[8pt] a &= -16 \\[8pt] \Rightarrow (4-a) &= 4-(-16) = 20 \end{aligned}
Jawaban E.
Contoh 20: SPMB 2005
Nilai \( \displaystyle \lim_{x \to 0} \ \frac{5x^2+x}{\sqrt{4+x}-2} = \cdots \)
- 0
- 1
- 2
- 4
- 6
\begin{aligned} \lim_{x \to 0} \ \frac{5x^2+x}{\sqrt{4+x}-2} &= \lim_{x \to 0} \ \frac{5x^2+x}{\sqrt{4+x}-2} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2} \\[8pt] &= \lim_{x \to 0} \ \frac{(5x^2+x)(\sqrt{4+x}+2)}{(4+x)-4} \\[8pt] &= \lim_{x \to 0} \ \frac{x(5x+1)(\sqrt{4+x}-2)}{x} \\[8pt] &= \lim_{x \to 0} \ (5x+1)(\sqrt{4+x}+2) \\[8pt] &= (5(0)+1)(\sqrt{4+0}+2) \\[8pt] &= (0+1)(2+2) = 4 \end{aligned}
Jawaban D.
Contoh 21: SPMB 2005
Nilai \( \displaystyle \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} = \cdots \)
- -1
- 0
- 2
- 6
- 8
\begin{aligned} \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} &= \lim_{x \to 2} \ \frac{4-x^2}{3-\sqrt{x^2+5}} \times \frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}} \\[8pt] &= \lim_{x \to 2} \ \frac{(4-x^2)(3+\sqrt{x^2+5})}{9-(x^2+5)} \\[8pt] &= \lim_{x \to 2} \ \frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2} \\[8pt] &= \lim_{x \to 2} \ (3+\sqrt{x^2+5}) = 3+\sqrt{2^2+5} \\[8pt] &= 3 + 3 = 6 \end{aligned}
Jawaban D.
Contoh 22: UM UGM 2004
Nilai \( \displaystyle \lim_{x \to 3} \ \frac{9-x^2}{4-\sqrt{x^2+7}} = \cdots \)
- 0
- 1
- 6
- 8
- ∞
\begin{aligned} \lim_{x \to 3} \ \frac{9-x^2}{4-\sqrt{x^2+7}} &= \lim_{x \to 3} \ \frac{9-x^2}{4-\sqrt{x^2+7}} \times \frac{4+\sqrt{x^2+7}}{4+\sqrt{x^2+7}} \\[8pt] &= \lim_{x \to 3} \ \frac{(9-x^2)(4+\sqrt{x^2+7})}{16-(x^2+7)} \\[8pt] &= \lim_{x \to 3} \ \frac{(9-x^2)(4+\sqrt{x^2+7})}{9-x^2} \\[8pt] &= \lim_{x \to 3} \ (4+\sqrt{x^2+7}) \\[8pt] &= 4 + \sqrt{3^2+7} = 8 \end{aligned}
Jawaban D.
Contoh 23: SPMB 2005
Nilai \( \displaystyle \lim_{x \to q} \ \frac{x\sqrt{x}-q\sqrt{q}}{\sqrt{x}-\sqrt{q}} = \cdots \)
- \( 3 \sqrt{q} \)
- \( \sqrt{q} \)
- \( q \)
- \( q \sqrt{q} \)
- \( 3q \)
\begin{aligned} \lim_{x \to q} \ \frac{x\sqrt{x}-q\sqrt{q}}{\sqrt{x}-\sqrt{q}} &= \lim_{x \to q} \ \frac{x\sqrt{x}-q\sqrt{q}}{\sqrt{x}-\sqrt{q}} \times \frac{\sqrt{x}+\sqrt{q}}{\sqrt{x}+\sqrt{q}} \\[8pt] &= \lim_{x \to q} \ \frac{x^2+x\sqrt{qx}-q\sqrt{qx}-q^2}{x-q} \\[8pt] &= \lim_{x \to q} \ \frac{(x-q)(x+q)+\sqrt{qx}(x-q)}{x-q} \\[8pt] &= \lim_{x \to q} \ (x+q)+\sqrt{qx} \\[8pt] &= (q+q) + \sqrt{q^2} = 2q + q \\[8pt] &= 3q \end{aligned}
Jawaban E.
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Penulis: Tju Ji Long · Statistisi
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