Pembahasan Soal UN SMA Materi Limit

Dibawah ini merupakan jenis soal dengan materi limit yang sering keluar saat UN. Mudah-mudahan UN 2018/2019 akan keluar juga jenis soal limit seperti ini. Jadi silahkan pelajari pembahasan berikut.

Konsep dasar yang digunakan dalam perhitungan beberapa soal di bawah adalah:
$\lim_{x \to \infty }\sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+qx+r}=\frac{b-q}{2\sqrt{a}}$

Soal 1, UN SMA Tapel 2017/2018 Program Studi IPA
Nilai dari $\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-4x+1=$...
A. $-\frac{9}{4}$
B. $-\frac{1}{4}$
C. $\frac{1}{4}$
D. $\frac{5}{4}$
E. $\frac{9}{4}$

Pembahasan
$\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-4x+1$
$=\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-\left(4x-1 \right )$
$=\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-\sqrt{\left(4x-1 \right )^{2}}$

Catatan: $\left ( ax+b \right )^{2}=a^{2}x^{2}+2\times a\times b\times x+b^{2}$

$=\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-\sqrt{16x^{2}-8x+1}$
$=\frac{10-(-8)}{2\sqrt{16}}$
$=\frac{18}{2.4}$
$=\frac{18}{8}$
$=\frac{9}{4}$
Jawabannya E

Soal 2, UN SMA Tapel 2016/2017 Program Studi IPA
Nilai dari $\lim_{x \to \infty }\left ( 2x-\sqrt{4x^{2}+x+3} \right )$ adalah ...
A. $-\frac{1}{2}$
B. $-\frac{1}{4}$
C. (0)
D. $\frac{1}{4}$
E. $\frac{1}{2}$

Pembahasan
$\lim_{x \to \infty }\left ( 2x-\sqrt{4x^{2}+x+3} \right )$
$\lim_{x \to \infty }\left ( \sqrt{\left ( 2x \right )^{2}}-\sqrt{4x^{2}+x+3} \right )$
$\lim_{x \to \infty }\left ( \sqrt{4x^{2}}-\sqrt{4x^{2}+x+3} \right )$
$\lim_{x \to \infty }\left ( \sqrt{4x^{2}+0x+0}-\sqrt{4x^{2}+x+3} \right )$
$=\frac{0-1}{2\sqrt{4}}$
$=\frac{-1}{2.2}$
$=\frac{-1}{4}$
$=-\frac{1}{4}$
Jawabannya B

Soal 3, UN SMA Tapel 2014/2015 Program Studi IPA
Nilai $\lim_{x \to \infty }\sqrt{x^{2}-6x+9}-(x-2)=$ adalah...
A. -1
B. -2
C. -3
D. -4
E. -5

Pembahasan
$\lim_{x \to \infty }\sqrt{x^{2}-6x+9}-(x-2)=$
$=\lim_{x \to \infty }\sqrt{x^{2}-6x+9}-\sqrt{\left(x-2 \right )^{2}}$
$=\lim_{x \to \infty }\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-4x+4}$
$=\frac{-6-(-4)}{2\sqrt{1}}$
$=\frac{-6+4}{2.1}$
$=\frac{-2}{2}$
$=-1$
Jawabannya A

Soal 4, UN SMA Tapel 2014/2015 Program Studi IPA
Nilai $\lim_{x \to \infty }\sqrt{4x^{2}+4x-3}-(2x-5)=$ adalah...
A. -6
B. -4
C. -1
D. 4
E. 6

Pembahasan
$\lim_{x \to \infty }\sqrt{4x^{2}+4x-3}-(2x-5)=$
$=\lim_{x \to \infty }\sqrt{4x^{2}+4x-3}-\sqrt{(2x-5)^{2}}$
$=\lim_{x \to \infty }\sqrt{4x^{2}+4x-3}-\sqrt{4x^{2}-20x+25}$
$=\frac{4-(-20)}{2\sqrt{4}}$
$=\frac{4+20}{2.2}$
$=\frac{24}{4}$
$=6$
Jawabannya E

Pembahasan Soal UN SMA Materi Limit Trigonometri

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

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]
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{\left ( 1-\sqrt{x-3} \right )\left ( 1+\sqrt{x-3} \right )}$

Catatan: $(a-b)(a+b)=a^{2}-b^{2}$

$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1^{2}-\left ( \sqrt{x-3} \right )^{2}}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1-\left ( {x-3} \right )}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{1-x+3}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{4-x}$
$=\lim_{x\rightarrow 4}\frac{\left ( x^{2}-16 \right )\left ( 1+\sqrt{x-3} \right )}{4-x}$
$=\lim_{x\rightarrow 4}\frac{(x-4)(x+4)\left ( 1+\sqrt{x-3} \right )}{-(x-4)}$
$=\lim_{x\rightarrow 4}\frac{(x+4)\left ( 1+\sqrt{x-3} \right )}{-1}$
$=\frac{(4+4)\left ( 1+\sqrt{4-3} \right )}{-1}$
$=\frac{8\times2}{-1}=-16$
Jawabannya E

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

Pembahasan
$\lim_{x\rightarrow 0}\frac{1-cos\left ( 4x \right )}{2x.sin \left ( 4x \right )}$

Catatan: $cos\left ( 4x \right )=1-2sin^{2}\left ( 2x \right )$

$=\lim_{x\rightarrow 0}\frac{1-\left ( 1-2sin^{2}\left ( 2x \right ) \right )}{2x.sin \left ( 4x \right )}$
$=\lim_{x\rightarrow 0}\frac{1-1+2sin^{2}\left ( 2x \right )}{2x.sin \left ( 4x \right )}$
$=\lim_{x\rightarrow 0}\frac{2sin^{2}\left ( 2x \right )}{2x.sin \left ( 4x \right )}$
$=\frac{2}{2}\times \lim_{x\rightarrow 0}\frac{sin\left ( 2x \right )}{x}\times \lim_{x\rightarrow 0}\frac{sin\left ( 2x \right )}{sin \left ( 4x \right )}$
$=\frac{2}{2}\times\frac{2}{1} \times\frac{2}{4}=1$
Jawabannya A

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

Pembahasan

Catatan:$sin^{2}\left ( 2x \right )+cos^{2}\left ( 2x \right )=1\Leftrightarrow 1-cos^{2}\left ( 2x \right )=sin^{2}\left ( 2x \right )$

$\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{1-cos^{2}\left ( 2x \right )}$
$=\lim_{x\rightarrow 0}\frac{2x.tan \left ( x \right )}{sin^{2}\left ( 2x \right )}$
$=2\lim_{x\rightarrow 0}\frac{x}{sin\left ( 2x \right )}\lim_{x\rightarrow 0}\frac{tan\left ( x \right )}{sin\left ( 2x \right )}$
$=\frac{2.1}{2.2}=\frac{1}{2}$
Jawabannya D