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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. 2
b. 0
c. 1
d. -1
e. 3
jawaban B
pembahasan :
\(\lim\limits_{x\rightarrow\infty}\frac{3x^{2}-4x^+6}{4x^{3}-2x^{2}-3x}\)=
\(\lim\limits_{x\rightarrow\infty}\frac{\frac{3x^{2}}{x^{3}}-\frac{4x}{x^{3}}+\frac{6}{x^{3}}}{\frac{4x^{3}}{x^{3}}-\frac{2x^{2}}{x^{3}}-\frac{3x}{x^{3}}}\)=
\(\lim\limits_{x\rightarrow\infty}\frac{\frac{3}{x}-\frac{4}{x^{2}}+\frac{6}{x^{3}}}{4-\frac{2}{x}-\frac{3}{x^{2}}}\)=
\(\frac{0-0+0}{4-0-0}=\frac{0}{4}0\)
2. \(\lim\limits_{x\rightarrow\infty }\frac{2x^{2}-x+5}{x^{2}-3x+2}\)=
a. 0
b. 1
c. 3
d. 2
e. 5
jawaban D
pembahasan :
\(\lim\limits_{x\rightarrow\infty }\frac{2x^{2}-x+5}{x^{2}-3x+2}\)=
\(\lim\limits_{x\rightarrow\infty}\frac{\frac{2x^{2}}{x^{2}}-\frac{x}{x^{2}}+\frac{5}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{3x}{x^{2}}+\frac{2}{x^{2}}}\)=
\(\lim\limits_{x\rightarrow\infty }\frac{2-\frac{1}{x}+\frac{5}{x^{2}}}{1-\frac{3}{x}+\frac{2}{x^{2}}}\)=
\(\frac{2-0+0}{1-0+0}\)=
\(\frac{2}{1}\)=2
3. \(\lim\limits_{x\rightarrow\infty } \frac{7x^{2}+3x+2}{2x-9}\)=
a. \(\infty \)
b. 3
c. 1
d. 0
e. 2
jawaban A
pembahasan :
\(\lim\limits_{x\rightarrow\infty } \frac{7x^{2}+3x+2}{2x-9}\)=
\(\lim\limits_{x\rightarrow \infty }\frac{\frac{7x^{2}}{x^{2}}+\frac{3x}{x^{2}}+\frac{2}{x^{2}}}{\frac{2x}{x^{2}}-\frac{9}{x^{2}}}\)=
\(\frac{7+0+0}{0-0}\)=
\(\frac{7}{0}\)=\(\infty \)
4. \(\lim\limits_{x\rightarrow \infty }\frac{6x^{2}-x+1}{x^{2}-4x+7}\)=
a. 2
b. 4
c. 3
d. 6
e. 0
jawaban D
pembahasan :
\(\lim\limits_{x\rightarrow \infty }\frac{6x^{2}-x+1}{x^{2}-4x+7}\)=
\(\lim\limits_{x\rightarrow \infty }\frac{\frac{6x^{2}}{x^{2}}-\frac{x}{x^{2}}+\frac{1}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{4}{x^{2}}+\frac{7}{x^{2}}}\)=
\(\frac{6-0+1}{1-0+1}\)=
\(\frac{6}{1}\)=6
5. Hasil dari \(\lim\limits_{x\rightarrow \infty }\frac{4x^{3}+2x^{2}-5}{8x^{3}-x+2}\)= adalah...
a. \frac{1}{2}
b. 3
c. 2
d. \(\frac{3}{2}\
e. 1
jawaban A
pembahasan :
\(\lim\limits_{x\rightarrow \infty }\frac{4x^{3}+2x^{2}-5}{8x^{3}-x+2}\)=
\(\lim\limits_{x\rightarrow \infty }\frac{\frac{4x^{3}}{x^{3}}+\frac{2x^{2}}{x^{3}}-\frac{5}{x}}{\frac{8x^{3}}{x^{3}}-\frac{x}{x^{3}}+\frac{2}{x^{3}}}\)=
\(\frac{4+0-0}{8-0+0}\)=
\(\frac{4}{8}\)=
\(\frac{1}{2}\)
6. \(\lim\limits_{x\rightarrow\infty }\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3}\)=
a. 2
b. 1
c. 5
d. \(\infty \)
e. 0
jawaban D
pembahasan :
\(\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3}\)=
\(\lim\limits_{x\rightarrow\infty } \sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3}\times \frac{\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3}}{\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3}}\)=
\(\lim\limits_{x\rightarrow\infty } \frac{2X^{2}+3X-1-\left ( X^{2}-5X+3 \right )}{\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3}}\)=
\(\lim\limits_{x\rightarrow\infty }\frac{2X^{2}+3X-1-X^{2}-5X-3}{\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3}}\)=
\(\lim\limits_{x\rightarrow\infty }\frac{X^{2}+8X-4}{\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3}}\)=
\(\lim\limits_{x\rightarrow\infty }\frac{\frac{X^{2}}{X}+\frac{8X}{X}-\frac{4}{X}}{\sqrt{\frac{2X^{2}}{X^{2}}+\frac{3X}{X^{2}}-\frac{1}{X^{2}}}+\sqrt{\frac{X^{2}}{X^{2}}-\frac{5X}{X^{2}}+\frac{3}{X^{2}}}}\)=
\(\frac{\infty +8-0}{\sqrt{2+0-0}+\sqrt{1-0+0}}\)=
\(\frac{\infty }{\sqrt{2}+\sqrt{1}}\)=
\(\infty \)
7. \(\lim\limits_{x\rightarrow\infty } \frac{4x^{3}+2x+1}{5x^{3}+8x^{2}+6}\)=
a. 2
b. \(\frac{4}{5}\)
c. 0
d. 5
e. 1
jawaban B
pembahasan :
\(\lim\limits_{x\rightarrow\infty } \frac{4x^{3}+2x+1}{5x^{3}+8x^{2}+6}\)=
\(\lim\limits_{x\rightarrow\infty }\frac{\frac{4x^{3}}{x^{3}}+\frac{2x}{x^{3}}+\frac{1}{x}}{\frac{5x^{3}}{x^{3}}+\frac{8x^{2}}{x^{3}}+\frac{6}{x^{3}}}\)=
\(\frac{4+0+0}{5+0+0}\)= \(\frac{4}{5}\)
8. \(\lim\limits_{x\rightarrow \infty } \frac{x^{2}-x-6}{3x^{2}+x+6}\)=
a. \(\frac{2}{5}\)
b. \(\frac{1}{5}\)
c. 3
d. \(-\frac{1}{3}\)
e. \(\frac{1}{3}\)
jawaban E
pembhasan :
\(\lim\limits_{x\rightarrow \infty } \frac{x^{2}-x-6}{3x^{2}+x+6}\)=
\(\lim\limits_{x\rightarrow \infty } \frac{\frac{x^{2}}{x^{2}}-\frac{x}{x^{2}}-\frac{6}{x^{2}}}{\frac{3x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{6}{x^{2}}}\)=
\(\frac{1-0-0}{3+0+0}\)= \(\frac{1}{3}\)
9. \(\lim\limits_{x\rightarrow \infty }\sqrt{x+2}-\sqrt{x+1}\)=
a. 0
b. 2
c. \(\frac{1}{2}\)
d. 1
e. 3
jawaban A
pembahasan :
\(\lim\limits_{x\rightarrow \infty }\sqrt{x+2}-\sqrt{x+1}\)=
\(\lim\limits_{x\rightarrow \infty }\sqrt{x+2}-\sqrt{x+1}\times \frac{\sqrt{x+2}+\sqrt{x+1}}{\sqrt{x+2}+\sqrt{x+1}}\)=
\(\lim\limits_{x\rightarrow \infty } \frac{\left ( x+2 \right )-\left ( x+1 \right )}{\sqrt{x+2}+\sqrt{x+1}}\)=
\(\lim\limits_{x\rightarrow \infty }\frac{1}{\sqrt{x+2}+\sqrt{x+1}}\)=
\(\lim\limits_{x\rightarrow \infty }\frac{\frac{1}{x}}{\sqrt{1+\frac{2}{x}}+\sqrt{1+\frac{1}{x}}}\)=
\(\frac{0}{\sqrt{1+0}+\sqrt{1+0}}\)= \(\frac{0}{2}\)=0
10. \(\lim\limits_{x\rightarrow \infty } \frac{5x-4}{x^{2}+2x-6}\)=
a. 1
b. 3
c. 7
d. 0
e. 2
jawaban D
pembahasan :
\(\lim\limits_{x\rightarrow \infty } \frac{5x-4}{x^{2}+2x-6}\)=
\(\lim\limits_{x\rightarrow \infty }\frac{\frac{5x}{x^{2}}-\frac{4}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{2x}{x^{2}}-\frac{6}{x^{2}}}\)=
\(\frac{0-0}{1+0-6}\)= \(\frac{0}{1}\)=0
