Soal 1, UN SMA Tapel 2016/2017 Program Studi IPA
Nilai $\lim_{x\rightarrow 4}\frac{x^{2}-16}{1-\sqrt{x-3}}$ adalah ...
A. -16
B. -4
C. 4
D. 16
E. 32
Nilai $\lim_{x\rightarrow 4}\frac{x^{2}-16}{1-\sqrt{x-3}}$ adalah ...
A. -16
B. -4
C. 4
D. 16
E. 32
Pembahasan
$\lim_{x\rightarrow 4}\frac{x^{2}-16}{1-\sqrt{x-3}}$
$=\lim_{x\rightarrow 4}\frac{x^{2}-16}{1-\sqrt{x-3}}\times \frac{1+\sqrt{x-3}}{1+\sqrt{x-3}}$ [Kalikan dengan sekawannya]
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{\left ( 1-\sqrt{x-3} \right )\left ( 1+\sqrt{x-3} \right )}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1^{2}-\left ( \sqrt{x-3} \right )^{2}}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1-\left ( {x-3} \right )}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1-x+3}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{4-x}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{4-x}$
$=\lim_{x\rightarrow 4}\frac{(x-4)(x+4)\left ( 1+\sqrt{x-3} \right )}{-(x-4)}$
$=\lim_{x\rightarrow 4}\frac{(x+4)\left ( 1+\sqrt{x-3} \right )}{-1}$
$=\frac{(4+4)\left ( 1+\sqrt{4-3} \right )}{-1}$
$=\frac{8\times2}{-1}=-16$
Jawabannya E
$\lim_{x\rightarrow 4}\frac{x^{2}-16}{1-\sqrt{x-3}}$
$=\lim_{x\rightarrow 4}\frac{x^{2}-16}{1-\sqrt{x-3}}\times \frac{1+\sqrt{x-3}}{1+\sqrt{x-3}}$ [Kalikan dengan sekawannya]
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{\left ( 1-\sqrt{x-3} \right )\left ( 1+\sqrt{x-3} \right )}$
| Catatan: $(a-b)(a+b)=a^{2}-b^{2}$ |
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1-\left ( {x-3} \right )}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1-x+3}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{4-x}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{4-x}$
$=\lim_{x\rightarrow 4}\frac{(x-4)(x+4)\left ( 1+\sqrt{x-3} \right )}{-(x-4)}$
$=\lim_{x\rightarrow 4}\frac{(x+4)\left ( 1+\sqrt{x-3} \right )}{-1}$
$=\frac{(4+4)\left ( 1+\sqrt{4-3} \right )}{-1}$
$=\frac{8\times2}{-1}=-16$
Jawabannya E
Soal 2, UN SMA Tapel 2015/2016 Program Studi IPA
Nilai $\lim_{x\rightarrow 0}\frac{1-cos\left ( 4x \right )}{2x.sin \left ( 4x \right )}=$ ...
A. 1
B. $\frac{1}{2}$
C. 0
D. $-\frac{1}{2}$
E. -1
Nilai $\lim_{x\rightarrow 0}\frac{1-cos\left ( 4x \right )}{2x.sin \left ( 4x \right )}=$ ...
A. 1
B. $\frac{1}{2}$
C. 0
D. $-\frac{1}{2}$
E. -1
Pembahasan
$\lim_{x\rightarrow 0}\frac{1-cos\left ( 4x \right )}{2x.sin \left ( 4x \right )}$
$=\lim_{x\rightarrow 0}\frac{1-\left ( 1-2sin^{2}\left ( 2x \right ) \right )}{2x.sin \left ( 4x \right )}$
$=\lim_{x\rightarrow 0}\frac{1-1+2sin^{2}\left ( 2x \right )}{2x.sin \left ( 4x \right )}$
$=\lim_{x\rightarrow 0}\frac{2sin^{2}\left ( 2x \right )}{2x.sin \left ( 4x \right )}$
$=\frac{2}{2}\times \lim_{x\rightarrow 0}\frac{sin\left ( 2x \right )}{x}\times \lim_{x\rightarrow 0}\frac{sin\left ( 2x \right )}{sin \left ( 4x \right )}$
$=\frac{2}{2}\times\frac{2}{1} \times\frac{2}{4}=1$
Jawabannya A
$\lim_{x\rightarrow 0}\frac{1-cos\left ( 4x \right )}{2x.sin \left ( 4x \right )}$
| Catatan: $cos\left ( 4x \right )=1-2sin^{2}\left ( 2x \right )$ |
$=\lim_{x\rightarrow 0}\frac{1-1+2sin^{2}\left ( 2x \right )}{2x.sin \left ( 4x \right )}$
$=\lim_{x\rightarrow 0}\frac{2sin^{2}\left ( 2x \right )}{2x.sin \left ( 4x \right )}$
$=\frac{2}{2}\times \lim_{x\rightarrow 0}\frac{sin\left ( 2x \right )}{x}\times \lim_{x\rightarrow 0}\frac{sin\left ( 2x \right )}{sin \left ( 4x \right )}$
$=\frac{2}{2}\times\frac{2}{1} \times\frac{2}{4}=1$
Jawabannya A
Soal 3, UN SMA Tapel 2014/2015 Program Studi IPA
Nilai $\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{1-cos^{2}\left ( 2x \right )}$ adalah ...
A. -1
B. $-\frac{1}{2}$
C. 0
D. $\frac{1}{2}$
E. 1
Nilai $\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{1-cos^{2}\left ( 2x \right )}$ adalah ...
A. -1
B. $-\frac{1}{2}$
C. 0
D. $\frac{1}{2}$
E. 1
Pembahasan
$\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{1-cos^{2}\left ( 2x \right )}$
$=\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{sin^{2}\left ( 2x \right )}$
$=2\lim_{x\rightarrow 0}\frac{x}{sin\left ( 2x \right )}\lim_{x\rightarrow 0}\frac{tan\left ( x \right )}{sin\left ( 2x \right )}$
$=\frac{2.1}{2.2}=\frac{1}{2}$
Jawabannya D
| Catatan:$sin^{2}\left ( 2x \right )+cos^{2}\left ( 2x \right )=1\Leftrightarrow 1-cos^{2}\left ( 2x \right )=sin^{2}\left ( 2x \right )$ |
$=\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{sin^{2}\left ( 2x \right )}$
$=2\lim_{x\rightarrow 0}\frac{x}{sin\left ( 2x \right )}\lim_{x\rightarrow 0}\frac{tan\left ( x \right )}{sin\left ( 2x \right )}$
$=\frac{2.1}{2.2}=\frac{1}{2}$
Jawabannya D
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