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Selamat datang di halaman contoh soal dan pembahasan "Limit".
Di halaman ini akan membahas tentang contoh soal dan pembahasan lengkap mengenai Limit.
Untuk lebih jelasnya mari kita lihat contoh soal dan pembahasan dibawah ini!
a. \(\frac{1}{16}\)
b. \(\frac{1}{32}\)
c. \(\frac{-1}{12}\)
d. \(\frac{-1}{32}\)
e. \(\frac{-1}{12}\)
jawaban B
\(\lim\limits_{x\rightarrow 4}\frac{\sqrt{x}-2}{x^2-16}\)=
Pembahasan
=\(\lim\limits_{x\rightarrow 4}\frac{-\frac{1}{2}x^{\frac{1}{2}}}{2x}\)
=\(\frac{\frac{1}{4}}{8}\)
=\(\frac{1}{32}\)
2. \(\lim\limits_{x\rightarrow 3}\frac{2- \sqrt {x+1}}{x-3}\)=
a. \(\frac{1}{3}\)
b. \(\frac{1}{2}\)
c. \(\frac{-1}{4}\)
d. \(\frac{-1}{2}\)
e. \(\frac{1}{12}\)
jawaban C
Pembahasan
=\(\lim\limits_{x\rightarrow 3}\frac{2-\left ( x+1 \right )^{}\frac{1}{2}}{x-3}\)
=\(\lim\limits_{x\rightarrow 3}\frac{-\frac{1}{2}\left ( x+1 \right )^\frac{1}{2}}{1}\)
=\(-\frac{1}{2}\left ( 4 \right )^\frac{1}{2}\)
=\(-\frac{1}{2\sqrt4}\)
=\(-\frac{1}{4}\)
3. \(\lim\limits_{x\rightarrow \infty }{(3x-2)-\sqrt{9x^2-2x+5}}\)=
a. \(\frac{1}{3}\)
b. \(\frac{-1}{2}\)
c. \(\frac{-1}{4}\)
d. \(\frac{-5}{3}\)
e. \(\frac{-1}{12}\)
jawaban D
\(\lim\limits_{x\rightarrow \infty }{(3x-2)-\sqrt{9x^2-2x+5}}\)=
Pembahasan
\(\lim\limits_{x\rightarrow \infty }\sqrt{9x^2-12x+4}-{\sqrt{9x^2-2x+5}}\)
=\(\frac{b-q}{2\sqrt{a}}=\frac{-12-\left ( 2 \right )}{2\sqrt{9}}\)
=\(-\frac{10}{6}\)
=\(-\frac{5}{3}\)
4.\(\lim\limits_{x\rightarrow \infty }\sqrt{9x^2+3x}-\sqrt{}{9x^2-5x}\)=
a. \(\frac{1}{3}\)
b. \(\frac{-1}{2}\)
c. \(\frac{-1}{4}\)
d. \(\frac{4}{3}\)
e. \(\frac{-1}{12}\)
jawaban D
Pembahasan
\(\frac{b-q}{2\sqrt{a}}=\frac{3-\left ( -5 \right )}{2\sqrt{9}}\)
=\(\frac{8}{6}\)
=\(\frac{4}{3}\)
5.\(\lim\limits_{x\rightarrow \infty }\sqrt{4x^2+5x}-\sqrt{4x^2-3}\)=
a. \(\frac{1}{3}\)
b. \(\frac{5}{4}\)
c. \(\frac{-1}{4}\)
d. \(\frac{-5}{3}\)
e. \(\frac{-1}{12}\)
jawaban B
Pembahasan
=\(\frac{b-q}{2\sqrt{a}}\)
=\(\frac{5-0}{2\sqrt{4}}\)
=\(\frac{5}{4}\)
6.\( \lim\limits_{x\rightarrow 1}\frac{\sqrt{x^2+3}+2}{x+1}\)=
a. 6
b. 2
c. 4
d. 10
e. 8
jawaban B
Pembahasan
=\(\frac{\sqrt{1^2+3}+2}{1+1}\)
=\(\frac{\sqrt{4}+2}{2}=2\)
7.\(\lim\limits_{x\rightarrow 4}\frac{3-\sqrt{x^2-7}}{x^2-2x-8}\)=
a. \(\frac{2}{9}\)
b. \(\frac{5}{4}\)
c. \(\frac{-1}{4}\)
d. \(\frac{-5}{3}\)
e. \(\frac{-1}{12}\)
jawaban A
pembahasan
=\(\lim\limits_{x\rightarrow 4}\frac{3-\left ( x^2-7 \right )^\frac{1}{2}}{x^2-2x-8}\)
=\(\frac{-\frac{4}{\sqrt{9}}}{6}\)
=\(\frac{\frac{4}{3}}{6}-\frac{4}{18}\)
=\(\frac{2}{9}\)
8. \(\lim\limits_{x\rightarrow 3}\frac{2- \sqrt {x+1}}{x-3}\)=
a. \(\frac{1}{3}\)
b. \(\frac{1}{2}\)
c. \(\frac{-1}{4}\)
d. \(\frac{-1}{2}\)
e. \(\frac{1}{12}\)
jawaban C
pembahasan
=\(\lim\limits_{x\rightarrow 3}\frac{2-\left ( x+1 \right )^{}\frac{1}{2}}{x-3}\)
=\(\lim\limits_{x\rightarrow 3}\frac{-\frac{1}{2}\left ( x+1 \right )^\frac{1}{2}}{1}\)
=\(-\frac{1}{2}\left ( 4 \right )^\frac{1}{2}\)
=\(-\frac{1}{2\sqrt4}\)
=\(-\frac{1}{4}\)
9. \(\lim\limits_{x\rightarrow 4}\frac{x-4}{\sqrt{x-2}}\)=
a. 6
b. 2
c. 4
d. 10
e. 8
jawaban C
pembahasan
=\(\lim\limits_{x\rightarrow 4}\frac{\sqrt{x+2}\sqrt{x-2}}{\sqrt{x-2}}\)
=\(\lim\limits_{x\rightarrow 4}\sqrt{x}+2\)
=\(\sqrt{4}+2=4\)
10. \(\lim\limits_{x\rightarrow \infty }\sqrt{25x^2-9x-6}-5x+3\) =
a. \(\frac{21}{10}\)
b. \(\frac{10}{2}\)
c. \(\frac{-10}{4}\)
d. \(\frac{-1}{21}\)
e. \(\frac{1}{12}\)
jawaban A
pembahasan
=\(\lim\limits_{x\rightarrow \infty }\sqrt{25x^2-9x-6}-\sqrt{\left ( 5x-3 \right )^2}\)
=\(\lim\limits_{x\rightarrow \infty }\sqrt{25x^2-9x-6-}\sqrt{25x^2-30x+9}\)
=\(\frac{-9-\left ( -30 \right )}{2\sqrt{25}}\)
=\(\frac{21}{10}\)
