Soal-Soal Matematika/Limit

Rumus umum

$\lim \limits _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$

Sifat limit

${\begin{matrix}\lim \limits _{x\to p}&k&=&k\\\lim \limits _{x\to p}&x&=&p\\\lim \limits _{x\to p}&f(x)&=&f(p)\\\lim \limits _{x\to p}&k\cdot f(x)&=&k\cdot \lim \limits _{x\to p}&f(x)\\\lim \limits _{x\to p}&(f(x)+g(x))&=&\lim \limits _{x\to p}f(x)+\lim \limits _{x\to p}g(x)\\\lim \limits _{x\to p}&(f(x)-g(x))&=&\lim \limits _{x\to p}f(x)-\lim \limits _{x\to p}g(x)\\\lim \limits _{x\to p}&(f(x)\cdot g(x))&=&\lim \limits _{x\to p}f(x)\cdot \lim \limits _{x\to p}g(x)\\\lim \limits _{x\to p}&(f(x)/g(x))&=&\lim \limits _{x\to p}f(x)/\lim \limits _{x\to p}g(x)\\\lim \limits _{x\to p}&(f(x))^{n}&=&(\lim \limits _{x\to p}f(x))^{n}\\\lim \limits _{x\to p}&{\sqrt[{n}]{f(x)}}&=&{\sqrt[{n}]{\lim \limits _{x\to p}f(x)}}\\\end{matrix}}$

Rumus lainnya

${\begin{matrix}\lim \limits _{x\to 0}&{\frac {x}{\sin x}}&=1\\\lim \limits _{x\to 0}&{\frac {x}{\tan x}}&=1\\\lim \limits _{x\to 0}&{\frac {\sin x}{x}}&=1\\\lim \limits _{x\to 0}&{\frac {\tan x}{x}}&=1\\\lim \limits _{x\to \infty }&x\sin({\frac {1}{x}})&=1\\\lim \limits _{x\to \infty }&x\tan({\frac {1}{x}})&=1\\\lim \limits _{x\to 0}&{\frac {ax}{\sin bx}}&={\frac {a}{b}}\\\lim \limits _{x\to 0}&{\frac {ax}{\tan bx}}&={\frac {a}{b}}\\\lim \limits _{x\to 0}&{\frac {\sin ax}{bx}}&={\frac {a}{b}}\\\lim \limits _{x\to 0}&{\frac {\tan ax}{bx}}&={\frac {a}{b}}\\\lim \limits _{x\to \infty }&p^{x}&=0,\qquad -1<p<1\\\lim \limits _{x\to \infty }&{\frac {ax^{m}+b}{px^{n}+q}}&={\frac {a}{p}},\qquad m=n\\\lim \limits _{x\to 0}&{\frac {{\frac {a}{x^{m}}}+b}{{\frac {p}{x^{n}}}+q}}&={\frac {a}{p}},\qquad m=n\\\lim \limits _{x\to \infty }&{\sqrt {ax^{2}+bx+c}}-{\sqrt {px^{2}+qx+r}}&={\frac {b-q}{2{\sqrt {a}}}},\qquad a=p\\\lim \limits _{x\to \infty }&{\sqrt[{3}]{ax^{3}+bx^{2}+cx+d}}-{\sqrt[{3}]{px^{3}+qx^{2}+rx+s}}&={\frac {b-q}{3{\sqrt[{3}]{a^{2}}}}},\qquad a=p\\\lim \limits _{x\to \infty }&(1+{\frac {1}{x}})^{x}&=e\\\lim \limits _{x\to 0}&(1+x)^{\frac {1}{x}}&=e\\\lim \limits _{x\to \infty }&(1+{\frac {a}{x}})^{bx}&=e^{ab}\\\lim \limits _{x\to 0}&(1+ax)^{\frac {b}{x}}&=e^{ab}\\\end{matrix}}$

Aturan L'Hôpital

$\lim \limits _{x\to p}{\frac {f(x)}{g(x)}}=\lim _{x\to p}{\frac {f'(x)}{g'(x)}}$

contoh soal

1. tentukan nilai limit dari

• $\lim _{x\to 3}{\frac {x^{2}-5x+6}{x^{2}-9}}$
• $\lim _{x\to p}{\frac {x^{2}-(3+p)x+3p}{x^{2}+(4-p)x-4p}}$
• $\lim _{x\to 9}{\frac {{\sqrt {x}}-3}{x^{2}-10x+9}}$
• $\lim _{x\to 8}{\frac {x^{2}-10x+16}{{\sqrt[{3}]{x}}-2}}$
• $\lim _{x\to 1}{\frac {(x-1)^{2}}{{\sqrt[{3}]{x^{2}}}-2{\sqrt[{3}]{x}}+1}}$
• $\lim _{x\to 0}{\frac {1-cosx}{8x^{2}}}$
• $\lim _{x\to b}{\frac {xsinb-bsinx}{x-b}}$
• $\lim _{x\to \infty }{\frac {16x^{5}+12x+3}{4x^{5}+108x^{3}+67x}}$
• $\lim _{x\to \infty }{\sqrt {9x^{2}+11x+2}}-{\sqrt {9x^{2}+5x+3}}$
• $\lim _{x\to \infty }({\frac {n+1}{n-3}})^{\frac {3n-9}{4}}$
• $\lim _{x\to \infty }({\frac {n^{2}+4n+10}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+9}}$
• $\lim _{x\to 4}{\frac {x^{2}-5x+4}{x^{2}-x-12}}$ (aturan L'Hôpital)

Jawab

• $\lim _{x\to 3}{\frac {x^{2}-5x+6}{x^{2}-9}}=\lim _{x\to 3}{\frac {(x-3)(x-2)}{(x-3)(x+3)}}=\lim _{x\to 3}{\frac {x-2}{x+3}}={\frac {3-2}{3+3}}={\frac {1}{6}}$
• $\lim _{x\to p}{\frac {x^{2}-(3+p)x+3p}{x^{2}+(4-p)x-4p}}=\lim _{x\to p}{\frac {(x-3)(x-p)}{(x+4)(x-p)}}=\lim _{x\to p}{\frac {x-3}{x+4}}={\frac {p-3}{p+4}}$
• $\lim _{x\to 9}{\frac {{\sqrt {x}}-3}{x^{2}-10x+9}}=\lim _{x\to 9}{\frac {({\sqrt {x}}-3)({\sqrt {x}}+3)}{(x-1)(x-9)({\sqrt {x}}+3)}}=\lim _{x\to 9}{\frac {x-9}{(x-1)(x-9)({\sqrt {x}}+3)}}=\lim _{x\to 9}{\frac {1}{(x-1)({\sqrt {x}}+3)}}={\frac {1}{(9-1)({\sqrt {9}}+3)}}={\frac {1}{48}}$
• $\lim _{x\to 8}{\frac {x^{2}-10x+16}{{\sqrt[{3}]{x}}-2}}=\lim _{x\to 8}{\frac {(x-8)(x-2)({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4)}{({\sqrt[{3}]{x}}-2)({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4)}}=\lim _{x\to 8}{\frac {(x-8)(x-2)({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4)}{x-8}}=\lim _{x\to 8}(x-2)({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4)=(8-2)({\sqrt[{3}]{8^{2}}}+2{\sqrt[{3}]{8}}+4)=6(4+4+4)=72$
• $\lim _{x\to 1}{\frac {(x-1)^{2}}{{\sqrt[{3}]{x^{2}}}-2{\sqrt[{3}]{x}}+1}}=\lim _{x\to 1}{\frac {({\sqrt[{3}]{x}}^{3}-1^{3})^{2}}{{\sqrt[{3}]{x^{2}}}-2{\sqrt[{3}]{x}}+1}}=\lim _{x\to 1}{\frac {({\sqrt[{3}]{x}}^{3}-1^{3})^{2}}{({\sqrt[{3}]{x}}-1)^{2}}}=\lim _{x\to 1}{\frac {(({\sqrt[{3}]{x}}-1)({\sqrt[{3}]{x^{2}}}+{\sqrt[{3}]{x}}+1))^{2}}{({\sqrt[{3}]{x}}-1)^{2}}}=\lim _{x\to 1}{\frac {({\sqrt[{3}]{x}}-1)^{2}({\sqrt[{3}]{x^{2}}}+{\sqrt[{3}]{x}}+1)^{2}}{({\sqrt[{3}]{x}}-1)^{2}}}=\lim _{x\to 1}({\sqrt[{3}]{x^{2}}}+{\sqrt[{3}]{x}}+1)^{2}=({\sqrt[{3}]{1^{2}}}+{\sqrt[{3}]{1}}+1)^{2}=(1+1+1)^{2}=9$
• $\lim _{x\to 0}{\frac {1-cosx}{8x^{2}}}=\lim _{x\to 0}{\frac {2sin^{2}{\frac {x}{2}}}{8x^{2}}}={\frac {2sin{\frac {x}{2}}sin{\frac {x}{2}}}{8xx}}={\frac {2}{8}}\cdot {\frac {1}{2}}\cdot {\frac {1}{2}}={\frac {1}{16}}$
• $\lim _{x\to b}{\frac {xsinb-bsinx}{x-b}}=\lim _{x\to b}{\frac {xsinb-bsinb+bsinb-bsinx}{x-b}}=\lim _{x\to b}{\frac {(x-b)sinb+b(sinb-sinx)}{x-b}}=\lim _{x\to b}{\frac {(x-b)sinb}{x-b}}+\lim _{x\to b}{\frac {b(sinb-sinx)}{x-b}}=sinb+b\lim _{x\to b}{\frac {sinb-sinx}{x-b}}=sinb+b\lim _{x\to b}{\frac {2cos{\frac {b+x}{2}}sin{\frac {b-x}{2}}}{x-b}}=sinb+b\lim _{x\to b}{\frac {2cos{\frac {b+x}{2}}sin{\frac {b-x}{2}}}{-2{\frac {b-x}{2}}}}=sinb-b\lim _{x\to b}{\frac {cos{\frac {b+x}{2}}sin{\frac {b-x}{2}}}{\frac {b-x}{2}}}=sinb-b\lim _{x\to b}cos{\frac {b+x}{2}}\cdot \lim _{x\to b}{\frac {sin{\frac {b-x}{2}}}{\frac {b-x}{2}}}=sinb-bcosb\cdot 1=sinb-bcosb$
• $\lim _{x\to \infty }{\frac {16x^{5}+12x+3}{4x^{5}+108x^{3}+67x}}=\lim _{x\to \infty }{\frac {x^{5}(16+{\frac {12}{x^{4}}}+{\frac {3}{x^{5}}})}{x^{5}(4+{\frac {108}{x^{2}}}+{\frac {67}{x^{4}}})}}={\frac {16}{4}}=4$
• $\lim _{x\to \infty }{\sqrt {9x^{2}+11x+2}}-{\sqrt {9x^{2}+5x+3}}={\frac {11-5}{2{\sqrt {9}}}}={\frac {6}{2\cdot 3}}=1$
• $\lim _{x\to \infty }({\frac {n+1}{n-3}})^{\frac {3n-9}{4}}=\lim _{x\to \infty }({\frac {n-3+4}{n-3}})^{\frac {3n-9}{4}}=\lim _{x\to \infty }({\frac {n-3}{n-3}}+{\frac {4}{n-3}})^{\frac {3(n-3)}{4}}=\lim _{x\to \infty }(1+{\frac {1}{\frac {n-3}{4}}})^{\frac {3(n-3)}{4}}=(\lim _{x\to \infty }(1+{\frac {1}{\frac {n-3}{4}}})^{\frac {n-3}{4}})^{3}=e^{3}$
• $\lim _{x\to \infty }({\frac {n^{2}+4n+10}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+21}}=\lim _{x\to \infty }({\frac {n^{2}+8n-4n+7+3}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+9}}=\lim _{x\to \infty }({\frac {(n^{2}-4n+7)+(8n+3)}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+9}}=\lim _{x\to \infty }({\frac {n^{2}-4n+7}{n^{2}-4n+7}}+{\frac {8n+3}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+9}}=\lim _{x\to \infty }(1+{\frac {1}{\frac {n^{2}-4n+7}{8n+3}}})^{\frac {2(n^{2}-4n+7)}{3(8n+3)}}=(\lim _{x\to \infty }(1+{\frac {1}{\frac {n^{2}-4n+7}{8n+3}}})^{\frac {n^{2}-4n+7}{8n+3}})^{\frac {2}{3}}=e^{\frac {2}{3}}={\sqrt[{3}]{e^{2}}}$
• $\lim _{x\to 4}{\frac {x^{2}-5x+4}{x^{2}-x-12}}=\lim _{x\to 4}{\frac {2x-5}{2x-1}}={\frac {2(4)-5}{2(4)-1}}={\frac {3}{7}}$

1. tentukan nilai a dan b dari $\lim _{x\to 5}{\frac {x^{2}+(a+b)x-10}{x^{2}+(a+6)x-15b}}={\frac {7}{11}}$!

Jawaban

$${\begin{aligned}\lim _{x\to 5}{\frac {x^{2}+(a+b)x-10}{x^{2}+(a+6)x-15b}}&={\frac {7}{11}}\\x^{2}+(a+b)x-10&=0\\(5)^{2}+(a+b)5-10&=0\\25+5a+5b-10&=0\\5a+5b&=-15\\a+b&=-3\\\lim _{x\to 5}{\frac {2x+(a+b)}{2x+(a+6)}}&={\frac {7}{11}}\\{\frac {2x+(a+b)}{2x+(a+6)}}&={\frac {7}{11}}\\{\frac {2(5)+(a+b)}{2(5)+(a+6)}}&={\frac {7}{11}}\\{\frac {10+(a+b)}{10+(a+6)}}&={\frac {7}{11}}\\{\frac {a+b+10}{a+16}}&={\frac {7}{11}}\\11(a+b+10)&=7(a+16)\\11a+11b+110&=7a+112\\4a+11b&=2\\a+b&=-3\\a&=-3-b\\4a+11b&=2\\4(-3-b)+11b&=2\\-12-4b+11b&=2\\7b&=14\\b&=2\\a&=-3-b\\a&=-3-2\\a&=-5\\\end{aligned}}$$