Soal-Soal Matematika/Limit

Rumus umum

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

dengan bentuk tak tentu: ${\frac {0}{0}},0^{0},0\cdot \infty ,{\frac {\infty }{\infty }},\infty -\infty ,1^{\infty },\infty ^{0}$

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

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

Aturan L'Hôpital

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

Aturan limit mutlak

Jika $\lim \limits _{x\to p}{\frac {|x-p|}{x-p}}$ maka hasilnya tidak ada karena batasan limit harus ada + atau -.

Jika $\lim \limits _{x\to p^{+}}{\frac {|x-p|}{x-p}}$ atau $\lim \limits _{x\to p^{-}}{\frac {|x-p|}{x-p}}$ maka hasilnya ada bila limit berlaku.

contoh soal

1. tentukan nilai limit dari

• $\lim _{x\to 3}{\frac {x^{2}-5x+6}{x^{2}-9}}$
• $\lim _{x\to p}{\frac {x^{2}-(3+p)x+3p}{x^{2}+(4-p)x-4p}}$
• $\lim _{x\to 9}{\frac {{\sqrt {x}}-3}{x^{2}-10x+9}}$
• $\lim _{x\to 8}{\frac {x^{2}-10x+16}{{\sqrt[{3}]{x}}-2}}$
• $\lim _{x\to 4}{\frac {({\sqrt {x}}-2)(5x+4)}{x^{2}-5x+4}}$
• $\lim _{x\to 1}{\frac {(x-1)^{2}}{{\sqrt[{3}]{x^{2}}}-2{\sqrt[{3}]{x}}+1}}$
• $\lim _{x\to 3}{\frac {{\frac {1}{3}}-{\frac {1}{2x-3}}}{x-3}}$
• $\lim _{x\to 0}{\frac {1-cosx}{8x^{2}}}$
• $\lim _{x\to b}{\frac {xsinb-bsinx}{x-b}}$
• $\lim _{x\to \infty }{\frac {16x^{5}+12x+3}{4x^{5}+108x^{3}+67x}}$
• $\lim _{x\to \infty }{\frac {2^{5x+4}-5^{2x}}{2^{5x+3}+7^{x}}}$
• $\lim _{x\to 0}{\frac {4e^{3x}-5(2)^{9x}+1}{3x}}$
• $\lim _{x\to 0}{\frac {9\cdot 15^{3x}-3^{3x+2}}{27x}}$
• $\lim _{x\to \infty }{\sqrt {9x^{2}+12x+7}}+{\sqrt {4x^{2}+8x+3}}-5x$
• $\lim _{x\to \infty }({\frac {n+1}{n-3}})^{\frac {3n-9}{4}}$
• $\lim _{x\to \infty }({\frac {n^{2}+4n+10}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+9}}$
• $\lim _{x\to \infty }({\frac {3n+8}{3n+7}})^{6n-5}$
• $\lim _{x\to \infty }({\frac {4n+3}{3n+1}})^{\frac {3n+1}{2n+5}}$
• $\lim _{x\to \infty }{\frac {5n^{2}}{1+3+5+\dots +n}}$
• $\lim _{x\to 5^{+}}{\frac {x^{2}-x+20}{|x-5|}}$
• $\lim _{x\to -6^{-}}{\frac {|x+6|}{x^{2}+2x-24}}$
• $\lim _{x\to 0}{\frac {40^{2x}-8^{2x}-5^{2x}+1}{4x^{2}}}$
• $\lim _{x\to 4}{\frac {x^{2}-5x+4}{x^{2}-x-12}}$ (aturan L'Hôpital)

Jawab

• $\lim _{x\to 3}{\frac {x^{2}-5x+6}{x^{2}-9}}=\lim _{x\to 3}{\frac {(x-3)(x-2)}{(x-3)(x+3)}}=\lim _{x\to 3}{\frac {x-2}{x+3}}={\frac {3-2}{3+3}}={\frac {1}{6}}$
• $\lim _{x\to p}{\frac {x^{2}-(3+p)x+3p}{x^{2}+(4-p)x-4p}}=\lim _{x\to p}{\frac {(x-3)(x-p)}{(x+4)(x-p)}}=\lim _{x\to p}{\frac {x-3}{x+4}}={\frac {p-3}{p+4}}$
• $\lim _{x\to 9}{\frac {{\sqrt {x}}-3}{x^{2}-10x+9}}=\lim _{x\to 9}{\frac {({\sqrt {x}}-3)({\sqrt {x}}+3)}{(x-1)(x-9)({\sqrt {x}}+3)}}=\lim _{x\to 9}{\frac {x-9}{(x-1)(x-9)({\sqrt {x}}+3)}}=\lim _{x\to 9}{\frac {1}{(x-1)({\sqrt {x}}+3)}}={\frac {1}{(9-1)({\sqrt {9}}+3)}}={\frac {1}{48}}$
• $\lim _{x\to 8}{\frac {x^{2}-10x+16}{{\sqrt[{3}]{x}}-2}}=\lim _{x\to 8}{\frac {(x-8)(x-2)({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4)}{({\sqrt[{3}]{x}}-2)({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4)}}=\lim _{x\to 8}{\frac {(x-8)(x-2)({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4)}{x-8}}=\lim _{x\to 8}(x-2)({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4)=(8-2)({\sqrt[{3}]{8^{2}}}+2{\sqrt[{3}]{8}}+4)=6(4+4+4)=72$
• $\lim _{x\to 4}{\frac {({\sqrt {x}}-2)(5x+4)}{x^{2}-5x+4}}=\lim _{x\to 4}{\frac {({\sqrt {x}}-2)(5x+4)}{(x-4)(x-1)}}=\lim _{x\to 4}{\frac {({\sqrt {x}}-2)(5x+4)}{({\sqrt {x}}-2)({\sqrt {x}}+2)(x-1)}}=\lim _{x\to 4}{\frac {5x+4}{({\sqrt {x}}+2)(x-1)}}={\frac {5(4)+4}{({\sqrt {4}}+2)(4-1)}}=2$
• $\lim _{x\to 1}{\frac {(x-1)^{2}}{{\sqrt[{3}]{x^{2}}}-2{\sqrt[{3}]{x}}+1}}=\lim _{x\to 1}{\frac {({\sqrt[{3}]{x}}^{3}-1^{3})^{2}}{{\sqrt[{3}]{x^{2}}}-2{\sqrt[{3}]{x}}+1}}=\lim _{x\to 1}{\frac {({\sqrt[{3}]{x}}^{3}-1^{3})^{2}}{({\sqrt[{3}]{x}}-1)^{2}}}=\lim _{x\to 1}{\frac {(({\sqrt[{3}]{x}}-1)({\sqrt[{3}]{x^{2}}}+{\sqrt[{3}]{x}}+1))^{2}}{({\sqrt[{3}]{x}}-1)^{2}}}=\lim _{x\to 1}{\frac {({\sqrt[{3}]{x}}-1)^{2}({\sqrt[{3}]{x^{2}}}+{\sqrt[{3}]{x}}+1)^{2}}{({\sqrt[{3}]{x}}-1)^{2}}}=\lim _{x\to 1}({\sqrt[{3}]{x^{2}}}+{\sqrt[{3}]{x}}+1)^{2}=({\sqrt[{3}]{1^{2}}}+{\sqrt[{3}]{1}}+1)^{2}=(1+1+1)^{2}=9$
• $\lim _{x\to 3}{\frac {{\frac {1}{3}}-{\frac {1}{2x-3}}}{x-3}}=\lim _{x\to 3}{\frac {\frac {2x-3-3}{3(2x-3)}}{x-3}}=\lim _{x\to 3}{\frac {\frac {2x-6}{3(2x-3)}}{x-3}}=\lim _{x\to 3}{\frac {2(x-3)}{3(2x-3)(x-3)}}=\lim _{x\to 3}{\frac {2}{3(2x-3)}}={\frac {2}{3(2(3)-3)}}={\frac {2}{3(3)}}={\frac {2}{9}}$
• $\lim _{x\to 0}{\frac {1-cosx}{8x^{2}}}=\lim _{x\to 0}{\frac {2sin^{2}{\frac {x}{2}}}{8x^{2}}}={\frac {2sin{\frac {x}{2}}sin{\frac {x}{2}}}{8xx}}={\frac {2}{8}}\cdot {\frac {1}{2}}\cdot {\frac {1}{2}}={\frac {1}{16}}$
• $\lim _{x\to b}{\frac {xsinb-bsinx}{x-b}}=\lim _{x\to b}{\frac {xsinb-bsinb+bsinb-bsinx}{x-b}}=\lim _{x\to b}{\frac {(x-b)sinb+b(sinb-sinx)}{x-b}}=\lim _{x\to b}{\frac {(x-b)sinb}{x-b}}+\lim _{x\to b}{\frac {b(sinb-sinx)}{x-b}}=sinb+b\lim _{x\to b}{\frac {sinb-sinx}{x-b}}=sinb+b\lim _{x\to b}{\frac {2cos{\frac {b+x}{2}}sin{\frac {b-x}{2}}}{x-b}}=sinb+b\lim _{x\to b}{\frac {2cos{\frac {b+x}{2}}sin{\frac {b-x}{2}}}{-2{\frac {b-x}{2}}}}=sinb-b\lim _{x\to b}{\frac {cos{\frac {b+x}{2}}sin{\frac {b-x}{2}}}{\frac {b-x}{2}}}=sinb-b\lim _{x\to b}cos{\frac {b+x}{2}}\cdot \lim _{x\to b}{\frac {sin{\frac {b-x}{2}}}{\frac {b-x}{2}}}=sinb-bcosb\cdot 1=sinb-bcosb$
• $\lim _{x\to \infty }{\frac {16x^{5}+12x+3}{4x^{5}+108x^{3}+67x}}=\lim _{x\to \infty }{\frac {x^{5}(16+{\frac {12}{x^{4}}}+{\frac {3}{x^{5}}})}{x^{5}(4+{\frac {108}{x^{2}}}+{\frac {67}{x^{4}}})}}={\frac {16}{4}}=4$
• $\lim _{x\to \infty }{\frac {2^{5x+4}-5^{2x}}{2^{5x+3}+7^{x}}}=\lim _{x\to \infty }{\frac {2^{5x}\cdot 2^{4}-5^{2x}}{2^{5x}\cdot 2^{3}+7^{x}}}=\lim _{x\to \infty }{\frac {32^{x}\cdot 16-25^{x}}{32^{x}\cdot 8+7^{x}}}=\lim _{x\to \infty }{\frac {32^{x}(16-({\frac {25}{32}})^{x})}{32^{x}(8+({\frac {7}{32}})^{x})}}={\frac {16}{8}}=2$
• $\lim _{x\to 0}{\frac {4e^{3x}-5(2)^{9x}+1}{3x}}=\lim _{x\to 0}{\frac {4e^{3x}-5(2^{3})^{3x}-4+5}{3x}}=\lim _{x\to 0}{\frac {4e^{3x}-4-(5(8)^{3x}-5)}{3x}}=\lim _{x\to 0}{\frac {4e^{3x}-4}{3x}}-\lim _{x\to 0}{\frac {5(8)^{3x}-5}{3x}}=4\lim _{x\to 0}{\frac {e^{3x}-1}{3x}}-5\lim _{x\to 0}{\frac {8^{3x}-1}{3x}}=4-5ln8$
• $\lim _{x\to 0}{\frac {9\cdot 15^{3x}-3^{3x+2}}{27x}}=\lim _{x\to 0}{\frac {9\cdot (3\cdot 5)^{3x}-3^{3x}\cdot 3^{2}}{27x}}=\lim _{x\to 0}{\frac {9\cdot 3^{3x}\cdot 5^{3x}-3^{3x}\cdot 9}{27x}}=\lim _{x\to 0}{\frac {9\cdot 3^{3x}\cdot (5^{3x}-1)}{27x}}=\lim _{x\to 0}{\frac {9\cdot 3^{3x}\cdot 5^{3x}-3^{3x}\cdot 9}{27x}}=\lim _{x\to 0}{\frac {9\cdot 3^{3x}\cdot (5^{3x}-1)}{9\cdot 3x}}=3^{0}ln5=ln5$
• $\lim _{x\to \infty }{\sqrt {9x^{2}+12x+7}}+{\sqrt {4x^{2}+8x+3}}-5x=\lim _{x\to \infty }{\sqrt {9x^{2}+12x+7}}+{\sqrt {4x^{2}+8x+3}}-3x-2x=\lim _{x\to \infty }({\sqrt {9x^{2}+12x+7}}-3x)+({\sqrt {4x^{2}+8x+3}}-2x)=\lim _{x\to \infty }({\sqrt {9x^{2}+12x+7}}-3x)+\lim _{x\to \infty }({\sqrt {4x^{2}+8x+3}}-2x)=\lim _{x\to \infty }({\sqrt {9x^{2}+12x+7}}-{\sqrt {9x^{2}}})+\lim _{x\to \infty }({\sqrt {4x^{2}+8x+3}}-{\sqrt {4x^{2}}})={\frac {12-0}{2{\sqrt {9}}}}+{\frac {8-0}{2{\sqrt {4}}}}=2+2=4$
• $\lim _{x\to \infty }({\frac {n+1}{n-3}})^{\frac {3n-9}{4}}=\lim _{x\to \infty }({\frac {n-3+4}{n-3}})^{\frac {3n-9}{4}}=\lim _{x\to \infty }({\frac {n-3}{n-3}}+{\frac {4}{n-3}})^{\frac {3(n-3)}{4}}=\lim _{x\to \infty }(1+{\frac {1}{\frac {n-3}{4}}})^{\frac {3(n-3)}{4}}=(\lim _{x\to \infty }(1+{\frac {1}{\frac {n-3}{4}}})^{\frac {n-3}{4}})^{3}=e^{3}$
• $\lim _{x\to \infty }({\frac {n^{2}+4n+10}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+21}}=\lim _{x\to \infty }({\frac {n^{2}+8n-4n+7+3}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+9}}=\lim _{x\to \infty }({\frac {(n^{2}-4n+7)+(8n+3)}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+9}}=\lim _{x\to \infty }({\frac {n^{2}-4n+7}{n^{2}-4n+7}}+{\frac {8n+3}{n^{2}-4n+7}})^{\frac {2n^{2}-8n+14}{24n+9}}=\lim _{x\to \infty }(1+{\frac {1}{\frac {n^{2}-4n+7}{8n+3}}})^{\frac {2(n^{2}-4n+7)}{3(8n+3)}}=(\lim _{x\to \infty }(1+{\frac {1}{\frac {n^{2}-4n+7}{8n+3}}})^{\frac {n^{2}-4n+7}{8n+3}})^{\frac {2}{3}}=e^{\frac {2}{3}}={\sqrt[{3}]{e^{2}}}$
• $\lim _{x\to \infty }({\frac {3n+8}{3n+7}})^{6n-5}=\lim _{x\to \infty }({\frac {3n+7+1}{3n+7}})^{6n-5}=\lim _{x\to \infty }({\frac {3n+7}{3n+7}}+{\frac {1}{3n+7}})^{6n-5}=\lim _{x\to \infty }(1+{\frac {1}{3n+7}})^{6n-5}=\lim _{x\to \infty }(1+{\frac {1}{3n+7}})^{\frac {(3n+7)(6n-5)}{3n+7}}=(\lim _{x\to \infty }(1+{\frac {1}{3n+7}})^{3n+7})^{\frac {6n-5}{3n+7}}=(\lim _{x\to \infty }(1+{\frac {1}{3n+7}})^{3n+7})^{\lim _{x\to \infty }{\frac {6n-5}{3n+7}}}=(\lim _{x\to \infty }(1+{\frac {1}{3n+7}})^{3n+7})^{\lim _{x\to \infty }{\frac {n(6-{\frac {5}{n}})}{n(3+{\frac {7}{n}})}}}=e^{\frac {6}{3}}=e^{2}$
• $\lim _{x\to \infty }({\frac {4n+3}{3n+1}})^{\frac {3n+1}{2n+5}}=\lim _{x\to \infty }({\frac {3n+1+n+2}{3n+1}})^{\frac {3n+1}{2n+5}}=\lim _{x\to \infty }({\frac {3n+1}{3n+1}}+{\frac {n+2}{3n+1}})^{\frac {3n+1}{2n+5}}=\lim _{x\to \infty }(1+{\frac {n+2}{3n+1}})^{\frac {3n+1}{2n+5}}=\lim _{x\to \infty }(1+{\frac {1}{\frac {3n+1}{n+2}}})^{\frac {3n+1}{2n+5}}=\lim _{x\to \infty }(1+{\frac {1}{\frac {3n+1}{n+2}}})^{{\frac {3n+1}{n+2}}\cdot {\frac {n+2}{2n+5}}}=(\lim _{x\to \infty }(1+{\frac {1}{\frac {3n+1}{n+2}}})^{\frac {3n+1}{n+2}})^{\frac {n+2}{2n+5}}=(\lim _{x\to \infty }(1+{\frac {1}{\frac {3n+1}{n+2}}})^{\frac {3n+1}{n+2}})^{\lim _{x\to \infty }{\frac {n+2}{2n+5}}}=(\lim _{x\to \infty }(1+{\frac {1}{\frac {3n+1}{n+2}}})^{\frac {3n+1}{n+2}})^{\lim _{x\to \infty }{\frac {n(1+{\frac {2}{n}})}{n(2+{\frac {5}{n}})}}}=e^{\frac {1}{2}}={\sqrt {e}}$
• $\lim _{x\to \infty }{\frac {5n^{2}}{1+3+5+\dots +n}}=\lim _{x\to \infty }{\frac {5n^{2}}{n^{2}}}=5$
• $\lim _{x\to 5^{+}}{\frac {x^{2}-x+20}{|x-5|}}=\lim _{x\to 5^{+}}{\frac {(x-5)(x+4)}{(x-5)}}=\lim _{x\to 5^{+}}x+4=5+4=9$
• $\lim _{x\to -6^{-}}{\frac {|x+6|}{x^{2}+2x-24}}=\lim _{x\to -6^{-}}{\frac {-(x+6)}{(x+6)(x-4)}}=\lim _{x\to -6^{-}}{\frac {-1}{x-4}}={\frac {-1}{-6-4}}={\frac {-1}{-10}}={\frac {1}{10}}$
• $\lim _{x\to 0}{\frac {40^{2x}-8^{2x}-5^{2x}+1}{4x^{2}}}=\lim _{x\to 0}{\frac {8^{2x}(5^{2x}-1)-(5^{2x}-1)}{4x^{2}}}=\lim _{x\to 0}{\frac {(8^{2x}-1)(5^{2x}-1)}{4x^{2}}}=\lim _{x\to 0}{\frac {(8^{2x}-1)(5^{2x}-1)}{2x\cdot 2x}}=ln8\cdot ln5=ln40$
• $\lim _{x\to 4}{\frac {x^{2}-5x+4}{x^{2}-x-12}}=\lim _{x\to 4}{\frac {2x-5}{2x-1}}={\frac {2(4)-5}{2(4)-1}}={\frac {3}{7}}$

alternatif

• $\lim _{x\to \infty }{\sqrt {9x^{2}+12x+7}}+{\sqrt {4x^{2}+8x+3}}-5x$
  • $\lim _{x\to \infty }{\sqrt {9x^{2}+12x+7}}+{\sqrt {4x^{2}+8x+3}}-5x$
  • $\lim _{x\to \infty }{\sqrt {(3x+2)^{2}}}+{\sqrt {(2x+2)^{2}}}-5x$
  • $\lim _{x\to \infty }3x+2+2x+2-5x$
  • 4

1. tentukan hasil dari $\lim _{x\to 0}({\frac {sinx}{x}})^{\frac {1}{x}}$!

Jawaban

$${\begin{aligned}\lim _{x\to 0}({\frac {sinx}{x}})^{\frac {1}{x}}\\\lim _{x\to 0}(1+{\frac {sinx}{x}}-1)^{\frac {1}{x}}\\\lim _{x\to 0}(1+{\frac {sinx-x}{x}})^{\frac {1}{x}}\\\lim _{x\to 0}(1+{\frac {sinx-x}{x}})^{({{\frac {1}{\frac {sinx-x}{x}}}\cdot {\frac {sinx-x}{x}}\cdot {\frac {1}{x}}})}\\(\lim _{x\to 0}(1+{\frac {sinx-x}{x}})^{\frac {1}{\frac {sinx-x}{x}}})^{({\frac {sinx-x}{x}}\cdot {\frac {1}{x}})}\\e^{\lim _{x\to 0}({\frac {sinx-x}{x}}\cdot {\frac {1}{x}})}\\e^{\lim _{x\to 0}{\frac {sinx-x}{x^{2}}}}\\{\text{misalkan }}p&=\lim _{x\to 0}{\frac {sinx-x}{x^{2}}}\\p&=\lim _{x\to 0}{\frac {sinx-x}{x^{2}}}\\{\text{ubah x menjadi 3x}}\\p&=\lim _{x\to 0}{\frac {sin3x-3x}{(3x)^{2}}}\\p&=\lim _{x\to 0}{\frac {3sinx-4sin^{3}x-3x}{9x^{2}}}\\p&=\lim _{x\to 0}{\frac {3(sinx-x)-4sin^{3}x}{9x^{2}}}\\p&=\lim _{x\to 0}{\frac {3(sinx-x)}{9x^{2}}}-\lim _{x\to 0}{\frac {4sin^{3}x}{9x^{2}}}\\p&={\frac {3}{9}}\lim _{x\to 0}{\frac {sinx-x}{x^{2}}}-\lim _{x\to 0}{\frac {4\cdot sinx\cdot sin^{2}x}{9x^{2}}}\\p&={\frac {1}{3}}p-0\\{\frac {2}{3}}p&=0\\p&=0\\\lim _{x\to 0}{\frac {sinx-x}{x^{2}}}&=0\\e^{\lim _{x\to 0}{\frac {sinx-x}{x^{2}}}}\\e^{0}\\1\\\lim _{x\to 0}({\frac {sinx}{x}})^{\frac {1}{x}}&=1\\\end{aligned}}$$

1. tentukan hasil dari $\lim _{x\to 0}({\frac {tanx}{x}})^{\frac {1}{x^{2}}}$!

Jawaban

$${\begin{aligned}\lim _{x\to 0}({\frac {tanx}{x}})^{\frac {1}{x^{2}}}\\\lim _{x\to 0}(1+{\frac {tanx}{x}}-1)^{\frac {1}{x^{2}}}\\\lim _{x\to 0}(1+{\frac {tanx-x}{x}})^{\frac {1}{x^{2}}}\\\lim _{x\to 0}(1+{\frac {tanx-x}{x}})^{({{\frac {1}{\frac {tanx-x}{x}}}\cdot {\frac {tanx-x}{x}}\cdot {\frac {1}{x^{2}}}})}\\(\lim _{x\to 0}(1+{\frac {tanx-x}{x}})^{\frac {1}{\frac {tanx-x}{x}}})^{({\frac {tanx-x}{x}}\cdot {\frac {1}{x^{2}}})}\\e^{\lim _{x\to 0}({\frac {tanx-x}{x}}\cdot {\frac {1}{x^{2}}})}\\e^{\lim _{x\to 0}{\frac {tanx-x}{x^{3}}}}\\{\text{misalkan }}p&=\lim _{x\to 0}{\frac {tanx-x}{x^{3}}}\\p&=\lim _{x\to 0}{\frac {tanx-x}{x^{3}}}\\p&=\lim _{x\to 0}{\frac {sec^{2}x-1}{3x^{2}}}{\text{(gunakan aturan L'Hôpital) }}\\p&=\lim _{x\to 0}{\frac {tan^{2}x}{3x^{2}}}\\p&={\frac {1}{3}}\\\lim _{x\to 0}{\frac {tanx-x}{x^{3}}}&={\frac {1}{3}}\\e^{\lim _{x\to 0}{\frac {tanx-x}{x^{3}}}}\\e^{\frac {1}{3}}\\{\sqrt[{3}]{e}}\\\lim _{x\to 0}({\frac {tanx}{x}})^{\frac {1}{x^{2}}}&={\sqrt[{3}]{e}}\\\end{aligned}}$$

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

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