Konsep dasar yang digunakan dalam perhitungan beberapa soal di bawah adalah:
$\lim_{x \to \infty }\sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+qx+r}=\frac{b-q}{2\sqrt{a}}$
$\lim_{x \to \infty }\sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+qx+r}=\frac{b-q}{2\sqrt{a}}$
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. $\frac{9}{4}$
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
$\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-4x+1$
$=\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-\left(4x-1 \right )$
$=\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-\sqrt{\left(4x-1 \right )^{2}}$
$=\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-\sqrt{16x^{2}-8x+1}$
$=\frac{10-(-8)}{2\sqrt{16}}$
$=\frac{18}{2.4}$
$=\frac{18}{8}$
$=\frac{9}{4}$
Jawabannya E
$\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-4x+1$
$=\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-\left(4x-1 \right )$
$=\lim_{x \to \infty }\sqrt{16x^{2}+10x-3}-\sqrt{\left(4x-1 \right )^{2}}$
Catatan: $\left ( ax+b \right )^{2}=a^{2}x^{2}+2\times a\times b\times x+b^{2}$ |
$=\frac{10-(-8)}{2\sqrt{16}}$
$=\frac{18}{2.4}$
$=\frac{18}{8}$
$=\frac{9}{4}$
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. $\frac{1}{2}$
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
$\lim_{x \to \infty }\left ( 2x-\sqrt{4x^{2}+x+3} \right )$
$\lim_{x \to \infty }\left ( \sqrt{\left ( 2x \right )^{2}}-\sqrt{4x^{2}+x+3} \right )$
$\lim_{x \to \infty }\left ( \sqrt{4x^{2}}-\sqrt{4x^{2}+x+3} \right )$
$\lim_{x \to \infty }\left ( \sqrt{4x^{2}+0x+0}-\sqrt{4x^{2}+x+3} \right )$
$=\frac{0-1}{2\sqrt{4}}$
$=\frac{-1}{2.2}$
$=\frac{-1}{4}$
$=-\frac{1}{4}$
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
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
$\lim_{x \to \infty }\sqrt{x^{2}-6x+9}-(x-2)=$
$=\lim_{x \to \infty }\sqrt{x^{2}-6x+9}-\sqrt{\left(x-2 \right )^{2}}$
$=\lim_{x \to \infty }\sqrt{x^{2}-6x+9}-\sqrt{x^{2}-4x+4}$
$=\frac{-6-(-4)}{2\sqrt{1}}$
$=\frac{-6+4}{2.1}$
$=\frac{-2}{2}$
$=-1$
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
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
$\lim_{x \to \infty }\sqrt{4x^{2}+4x-3}-(2x-5)=$
$=\lim_{x \to \infty }\sqrt{4x^{2}+4x-3}-\sqrt{(2x-5)^{2}}$
$=\lim_{x \to \infty }\sqrt{4x^{2}+4x-3}-\sqrt{4x^{2}-20x+25}$
$=\frac{4-(-20)}{2\sqrt{4}}$
$=\frac{4+20}{2.2}$
$=\frac{24}{4}$
$=6$
Jawabannya E
No comments:
Post a Comment