Pembahasan soal-soal Ujian Nasional (UN) SMA bidang studi Matematika IPA untuk pokok bahasan Limit Fungsi Aljabar.
1. EBT 2003
Nilai dari \(\mathrm{_{x \to 2}^{lim}\frac{4-x^{2}}{3-\sqrt{x^{2}+5}}}\) = …
A. −12
B. −6
C. 0
D. 6
E. 12
Pembahasan :
\(\begin{align} \mathrm{\lim_{x \to 2}\,\frac{4-x^{2}}{3-\sqrt{x^{2}+5}}} & = \mathrm{\lim_{x \to 2}\,\frac{4-x^{2}}{3-\sqrt{x^{2}+5}}\cdot \frac{3+\sqrt{x^{2}+5}}{3+\sqrt{x^{2}+5}}} \\ & = \mathrm{\lim_{x \to 2}\,\frac{\left ( 4-x^{2} \right )\left (3+\sqrt{x^{2}+5} \right )}{9-\left (x^{2}+5 \right )}} \\ & = \mathrm{\lim_{x \to 2}\,\frac{\left ( 4-x^{2} \right )\left (3+\sqrt{x^{2}+5} \right )}{4-x^{2}}} \\ & = \mathrm{\lim_{x \to 2}\,\left ( 3+\sqrt{x^{2}+5} \right )} \\ & = 3+\sqrt{2^{2}+5} \\ & = 6 \end{align}\)
Jawaban : D
2. UN 2004
Nilai \(\mathrm{_{x \to 2}^{lim}\left (\frac{2}{x^{2}-4}-\frac{3}{x^{2}+2x-8} \right )}\) = …
A. \(-\frac{7}{12}\)
B. \(-\frac{1}{4}\)
C. \(-\frac{1}{12}\)
D. \(-\frac{1}{24}\)
E. 0
Pembahasan :
\(\begin{align}
& \mathrm{\lim_{x \to 2}\left (\frac{2}{x^{2}-4}-\frac{3}{x^{2}+2x-8} \right )} \\
& = \mathrm{\lim_{x \to 2}\left (\frac{2}{(x+2)(x-2)}-\frac{3}{(x+4)(x-2)} \right )} \\
& = \mathrm{\lim_{x \to 2}\,\frac{2(x+4)-3(x+2)}{(x+2)(x-2)(x+4)}} \\
& = \mathrm{\lim_{x \to 2}\,\frac{-(x-2)}{(x+2)(x-2)(x+4)}} \\
& = \mathrm{\lim_{x \to 2}\,\frac{-1}{(x+2)(x+4)}} \\
& = \frac{-1}{(2+2)(2+4)} \\
& = -\frac{1}{24}
\end{align}\)
Jawaban : D
3. UN 2006
Nilai \(\mathrm{_{x \to 6}^{lim}\frac{\sqrt{3x-2}-\sqrt{2x+4}}{x-6}}\) = …
A. \(-\frac{1}{4}\)
B. \(-\frac{1}{8}\)
C. 0
D. \(\frac{1}{8}\)
E. \(\frac{1}{4}\)
Pembahasan :
\(\begin{align}
& \mathrm{\lim_{x \to 6}\frac{\sqrt{3x-2}\,-\sqrt{2x+4}}{x-6}} \\
& = \mathrm{\lim_{x \to 6}\,\frac{\sqrt{3x-2}\,-\sqrt{2x+4}}{x-6}\cdot \frac{\sqrt{3x-2}\,+\sqrt{2x+4}}{\sqrt{3x-2}\,+\sqrt{2x+4}}} \\
& = \mathrm{\lim_{x \to 6}\,\frac{(3x-2)-(2x+4)}{(x-6)\left (\sqrt{3x-2}\,+\sqrt{2x+4} \right )}} \\
& = \mathrm{\lim_{x \to 6}\,\frac{(x-6)}{(x-6)\left (\sqrt{3x-2}\,+\sqrt{2x+4} \right )}} \\
& = \mathrm{\lim_{x \to 6}\,\frac{1}{\sqrt{3x-2}\,+\sqrt{2x+4}}} \\
& = \frac{1}{\sqrt{3(6)-2}\,+\sqrt{2(6)+4}} \\
& = \frac{1}{8}
\end{align}\)
Jawaban : D
4. UN 2007
Nilai \(\mathrm{_{x \to 3}^{lim}\frac{x^{2}-x-6}{4-\sqrt{5x+1}}}\) = …
A. −8
B. −6
C. 6
D. 8
E. ∞
Pembahasan :
\(\begin{align}
\mathrm{\lim_{x \to 3}\,\frac{x^{2}-x-6}{4-\sqrt{5x+1}}}
& = \mathrm{\lim_{x \to 3}\,\frac{x^{2}-x-6}{4-\sqrt{5x+1}}\cdot \frac{4+\sqrt{5x+1}}{4+\sqrt{5x+1}}} \\
& = \mathrm{\lim_{x \to 3}\,\frac{(x^{2}-x-6)\left ( 4+\sqrt{5x+1} \right )}{16-(5x+1)}} \\
& = \mathrm{\lim_{x \to 3}\,\frac{(x-3)(x+2)\left ( 4+\sqrt{5x+1} \right )}{-5(x-3)}} \\
& = \mathrm{\lim_{x \to 3}\,\frac{(x+2)\left ( 4+\sqrt{5x+1} \right )}{-5}} \\
& = \frac{(3+2)\left ( 4+\sqrt{5(3)+1} \right )}{-5} \\
& = -8
\end{align}\)
Jawaban : A
5. UN 2008
Nilai \(\mathrm{_{x \to 2}^{lim}\frac{x^{3}-4x}{x-2}}\) = …
A. 32
B. 16
C. 8
D. 4
E. 2
Pembahasan :
\(\begin{align}
\mathrm{\lim_{x \to 2}\,\frac{x^{3}-4x}{x-2}}
& = \mathrm{\lim_{x \to 2}\,\frac{x\left (x^{2}-4 \right )}{x-2}} \\
& = \mathrm{\lim_{x \to 2}\,\frac{x(x+2)(x-2)}{x-2}} \\
& = \mathrm{\lim_{x \to 2}\,x(x+2)} \\
& = 2(2+2) \\
& = 8
\end{align}\)
Jawaban : C
6. UN 2009
Nilai \(\mathrm{_{x \to 3}^{lim}\frac{x^{2}-9}{\sqrt{10+2x}-(x+1)}}\) = …
A. −8
B. −6
C. 4
D. 6
E. 8
Pembahasan :
\(\begin{align}
& \mathrm{\lim_{x \to 3}\,\frac{x^{2}-9}{\sqrt{10+2x}\,-(x+1)}} \\
& = \mathrm{\lim_{x \to 3}\,\frac{x^{2}-9}{\sqrt{10+2x}\,-(x+1)}\cdot \frac{\sqrt{10+2x}\,+(x+1)}{\sqrt{10+2x}\,+(x+1)}} \\
& = \mathrm{\lim_{x \to 3}\,\frac{\left (x^{2}-9 \right )\left ( \sqrt{10+2x}\,+(x+1) \right )}{10+2x-(x^{2}+2x+1)}} \\
& = \mathrm{\lim_{x \to 3}\,\frac{\left (x^{2}-9 \right )\left ( \sqrt{10+2x}\,+(x+1) \right )}{-\left (x^{2}-9 \right )}} \\
& = \mathrm{\lim_{x \to 3}\,\frac{\sqrt{10+2x}\,+(x+1)}{-1}} \\
& = \mathrm{\frac{ \sqrt{10+2(3)}\,+(3+1) }{-1}} \\
& = -8
\end{align}\)
Jawaban : A
7. UN 2010
Nilai \(\mathrm{_{x \to 0}^{lim}\left ( \frac{4x}{\sqrt{1-2x}-\sqrt{1+2x}} \right )}\) = …
A. −2
B. 0
C. 1
D. 2
E. 4
Pembahasan :
\(\begin{align}
& \mathrm{\lim_{x \to 0}\frac{4x}{\sqrt{1-2x}-\sqrt{1+2x}}} \\
& = \mathrm{\lim_{x \to 0}\frac{4x}{\sqrt{1-2x}-\sqrt{1+2x}}\cdot \frac{\sqrt{1-2x}+\sqrt{1+2x}}{\sqrt{1-2x}+\sqrt{1+2x}}} \\
& = \mathrm{\lim_{x \to 0}\frac{4x\left ( \sqrt{1-2x}+\sqrt{1+2x} \right )}{(1-2x)-(1+2x)}} \\
& = \mathrm{\lim_{x \to 0}\frac{4x\left ( \sqrt{1-2x}+\sqrt{1+2x} \right )}{-4x}} \\
& = \mathrm{\lim_{x \to 0}\frac{\sqrt{1-2x}+\sqrt{1+2x} }{-1}} \\
& = \mathrm{\frac{\sqrt{1-2(0)}+\sqrt{1+2(0)} }{-1}} \\
& = -2
\end{align}\)
Jawaban : A
8. UN 2011
Nilai \(\mathrm{_{x \to 4}^{lim}\frac{(x-4)}{\sqrt{x}-2}}\) = …
A. 0
B. 4
C. 8
D. 12
E. 16
Pembahasan :
\(\begin{align}
\mathrm{\lim_{x \to 4}\frac{x-4}{\sqrt{x}-2}} & = \mathrm{\lim_{x \to 4}\frac{\left (\sqrt{x}-2 \right )\left ( \sqrt{x}+2 \right )}{\sqrt{x}-2}} \\
& = \mathrm{\lim_{x \to 4}\left ( \sqrt{x}+2 \right )} \\
& = \sqrt{4}+2 \\
& = 4
\end{align}\)
Jawaban : B
9. UN 2012
Nilai \(\mathrm{_{x \to 3}^{lim}\frac{2-\sqrt{x+1}}{x-3}=…}\)
A. \(-\frac{1}{4}\)
B. \(-\frac{1}{2}\)
C. 1
D. 2
E. 4
Pembahasan :
\(\begin{align}
\mathrm{\lim_{x \to 3}\frac{2-\sqrt{x+1}}{x-3}} & = \mathrm{\lim_{x \to 3}\frac{2-\sqrt{x+1}}{x-3}\cdot \frac{2+\sqrt{x+1}}{2+\sqrt{x+1}}} \\
& = \mathrm{\lim_{x \to 3}\frac{4-(x+1)}{(x-3)\left ( 2+\sqrt{x+1} \right )}} \\
& = \mathrm{\lim_{x \to 3}\frac{-(x-3)}{(x-3)\left ( 2+\sqrt{x+1} \right )}} \\
& = \mathrm{\lim_{x \to 3}\frac{-1}{2+\sqrt{x+1}}} \\
& = \frac{-1}{2+\sqrt{3+1}} \\
& = -\frac{1}{4}
\end{align}\)
Jawaban : A
10. UN 2013
Nilai dari \(\begin{align}
\mathrm{\lim_{x \to \infty }\left ( (2x-1)-\sqrt{4x^{2}-6x-5} \right )=…}
\end{align}\)
A. 4
B. 2
C. 1
D. \(\frac{1}{2}\)
E. \(\frac{1}{4}\)
Pembahasan :
Misalkan \(\begin{align}
\mathrm{\lim_{x \to \infty }\left ( (2x-1)-\sqrt{4x^{2}-6x-5} \right )=L}
\end{align}\)
\(\begin{align}
\mathrm{L}
& =\mathrm{\lim_{x \to \infty }\left ( \sqrt{(2x-1)^{2}}-\sqrt{4x^{2}-6x-5} \right )} \\
& = \mathrm{\lim_{x \to \infty }\left ( \sqrt{4x^{2}-4x+1}-\sqrt{4x^{2}-6x-5} \right )} \\
\end{align}\)
a = 4, b = -4, c = 1
p = 4, q = -6, r = -5
Karena a = p maka berlaku
\(\begin{align}
\mathrm{L = \frac{b-q}{2\sqrt{a}}=\frac{-4-(-6)}{2\sqrt{4}}=\frac{2}{4}=\frac{1}{2}}
\end{align}\)
Jawaban : D
11. UN 2013
Nilai dari \(\begin{align}
\mathrm{\lim_{x \to \infty }\;\frac{\sqrt{5-4x+3x^{2}}+\sqrt{4-3x+3x^{2}}}{2x}=…}
\end{align}\)
A. 0
B. \(\frac{1}{3}\)√3
C. √3
D. 2√3
E. ∞
Pembahasan :
Karena pangkat tertinggi pembilang sama dengan pangkat tertinggi penyebut, maka nilai limit tak hingga diatas sama dengan koefisien pangkat tertinggi pembilang dibagi koefisien pangkat tertinggi penyebut.
\(\begin{align}
\mathrm{\lim_{x \to \infty }\;\frac{\sqrt{5-4x+3x^{2}}+\sqrt{4-3x+3x^{2}}}{2x}} & =\frac{\sqrt{3}+\sqrt{3}}{2} \\
& = \sqrt{3}
\end{align}\)
Jawaban : C
12. UN 2014
Nilai \(\begin{align}
\mathrm{\lim_{x \to \infty }\left ( \sqrt{9x^{2}+6x-2}-3x+1 \right )=…}
\end{align}\)
A. 5
B. 4
C. 3
D. 2
E. 1
Pembahasan :
Misalkan \(\begin{align}
\mathrm{\lim_{x \to \infty }\left ( \sqrt{9x^{2}+6x-2}-3x+1 \right )=L}
\end{align}\)
\(\begin{align}
\mathrm{L}
& = \mathrm{\lim_{x \to \infty }\left ( \sqrt{9x^{2}+6x-2}-(3x-1) \right )} \\
& = \mathrm{\lim_{x \to \infty }\left ( \sqrt{9x^{2}+6x-2}-\sqrt{(3x-1)^{2}} \right )} \\
& = \mathrm{\lim_{x \to \infty }\left ( \sqrt{9x^{2}+6x-2}-\sqrt{9x^{2}-6x+1} \right )} \\
\end{align}\)
a = 9, b = 6, c = -2
p = 9, q = -6, r = 1
Karena a = p, maka berlaku :
\(\begin{align}
\mathrm{L=\frac{b-q}{2\sqrt{a}}=\frac{6-(-6)}{2\sqrt{9}}=\frac{12}{6}=2}
\end{align}\)
Jawaban : D
13. UN 2016
Nilai dari \(\begin{align}
\mathrm{\lim_{x \to \infty }\left ( \sqrt{4x^{2}+4x-3}-(2x-5) \right )=…}
\end{align}\)
A. −6
B. −4
C. −1
D. 4
E. 6
Pembahasan :
Misalkan \(\begin{align}
\mathrm{\lim_{x \to \infty }\left ( \sqrt{4x^{2}+4x-3}-(2x-5) \right )=L}
\end{align}\)
\(\begin{align}
\mathrm{L}
& = \mathrm{\lim_{x \to \infty }\left ( \sqrt{4x^{2}+4x-3}-\sqrt{(2x-5)^{2}} \right )} \\
& = \mathrm{\lim_{x \to \infty }\left ( \sqrt{4x^{2}+4x-3}-\sqrt{4x^{2}-20x+25} \right )} \\
\end{align}\)
a = 4, b = 4, c = -3
p = 4, q = -20, r = 25
Karena a = p maka berlaku
\(\begin{align}
\mathrm{L=\frac{b-q}{2\sqrt{a}}=\frac{4-(-20)}{2\sqrt{4}}=\frac{24}{4}=6}
\end{align}\)
Jawaban : E