lim h → 0 f ( x + h ) − f ( x ) h {\displaystyle \lim \limits _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}
lim x → p k = k lim x → p x = p lim x → p f ( x ) = f ( p ) lim x → p k ⋅ f ( x ) = k ⋅ lim x → p f ( x ) lim x → p ( f ( x ) + g ( x ) ) = lim x → p f ( x ) + lim x → p g ( x ) lim x → p ( f ( x ) − g ( x ) ) = lim x → p f ( x ) − lim x → p g ( x ) lim x → p ( f ( x ) ⋅ g ( x ) ) = lim x → p f ( x ) ⋅ lim x → p g ( x ) lim x → p ( f ( x ) / g ( x ) ) = lim x → p f ( x ) / lim x → p g ( x ) lim x → p ( f ( x ) ) n = ( lim x → p f ( x ) ) n lim x → p f ( x ) n = lim x → p f ( x ) n {\displaystyle {\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}}}
lim x → 0 x sin x = 1 lim x → 0 x tan x = 1 lim x → 0 sin x x = 1 lim x → 0 tan x x = 1 lim x → ∞ x sin ( 1 x ) = 1 lim x → ∞ x tan ( 1 x ) = 1 lim x → 0 a x sin b x = a b lim x → 0 a x tan b x = a b lim x → 0 sin a x b x = a b lim x → 0 tan a x b x = a b lim x → ∞ p x = 0 , − 1 < p < 1 lim x → ∞ a x m + b p x n + q = a p , m = n lim x → 0 a x m + b p x n + q = a p , m = n lim x → ∞ a x 2 + b x + c − p x 2 + q x + r = b − q 2 a , a = p lim x → ∞ a x 3 + b x 2 + c x + d 3 − p x 3 + q x 2 + r x + s 3 = b − q 3 a 2 3 , a = p lim x → ∞ ( 1 + 1 x ) x = e lim x → 0 ( 1 + x ) 1 x = e lim x → ∞ ( 1 + a x ) b x = e a b lim x → 0 ( 1 + a x ) b x = e a b lim x → 0 e x − 1 x = 1 lim x → 0 a x − 1 x = l n a {\displaystyle {\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 {e^{x}-1}{x}}&=1\\\lim \limits _{x\to 0}&{\frac {a^{x}-1}{x}}&=lna\\\end{matrix}}}
lim x → p f ( x ) g ( x ) = lim x → p f ′ ( x ) g ′ ( x ) {\displaystyle \lim \limits _{x\to p}{\frac {f(x)}{g(x)}}=\lim _{x\to p}{\frac {f'(x)}{g'(x)}}}
contoh soal
lim x → 5 x 2 + ( a + b ) x − 10 x 2 + ( a + 6 ) x − 15 b = 7 11 x 2 + ( a + b ) x − 10 = 0 ( 5 ) 2 + ( a + b ) 5 − 10 = 0 25 + 5 a + 5 b − 10 = 0 5 a + 5 b = − 15 a + b = − 3 lim x → 5 2 x + ( a + b ) 2 x + ( a + 6 ) = 7 11 2 x + ( a + b ) 2 x + ( a + 6 ) = 7 11 2 ( 5 ) + ( a + b ) 2 ( 5 ) + ( a + 6 ) = 7 11 10 + ( a + b ) 10 + ( a + 6 ) = 7 11 a + b + 10 a + 16 = 7 11 11 ( a + b + 10 ) = 7 ( a + 16 ) 11 a + 11 b + 110 = 7 a + 112 4 a + 11 b = 2 a + b = − 3 a = − 3 − b 4 a + 11 b = 2 4 ( − 3 − b ) + 11 b = 2 − 12 − 4 b + 11 b = 2 7 b = 14 b = 2 a = − 3 − b a = − 3 − 2 a = − 5 {\displaystyle {\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}}}
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4735793311d91e335a53e7677f0c0d05698df10)
![{\displaystyle {\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 {e^{x}-1}{x}}&=1\\\lim \limits _{x\to 0}&{\frac {a^{x}-1}{x}}&=lna\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f339a69107f317de5ce555e9fcad8503f90ebea8)
![{\displaystyle \lim _{x\to 8}{\frac {x^{2}-10x+16}{{\sqrt[{3}]{x}}-2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3523853cf48263524c6642fbf449b405fb251f8)
![{\displaystyle \lim _{x\to 1}{\frac {(x-1)^{2}}{{\sqrt[{3}]{x^{2}}}-2{\sqrt[{3}]{x}}+1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3608b8dba270c1c8bc6f8dfdb500b16c5f9212ea)
(aturan L'Hôpital)
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/266a87f87258c0a9532feae9577f9c1d72f77a1b)
![{\displaystyle \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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d96003fa33b67464462c9bfe2112234007fac3fe)
![{\displaystyle \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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d6ad5c7710e52c78365a8a66fa0687235602f1b)
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