Menyelesaikan limit dengan cara substitusi
Cara substitusi maksudnya langsung nilai $ x \, $ kita substitusi ke fungsi $ f(x) $. Contohnya :
$ \displaystyle \lim_{x \to a } f(x) = f(a) $
1). Tentukan nilai limit dari bentuk berikut :
a). $ \displaystyle \lim_{x \to 2 } 2x + 1 $
b). $ \displaystyle \lim_{x \to -1 } \frac{x^2 + 2}{2x - 1 } $
Penyelesaian :
a). $ \displaystyle \lim_{x \to 2 } 2x + 1 = 2(2) + 1 = 4 + 1 = 5 $
artinya nilai $ \displaystyle \lim_{x \to 2 } 2x + 1 = 5 $
b). $ \displaystyle \lim_{x \to -1 } \frac{x^2 + 2}{2x - 1 } = \frac{(-1)^2 + 2}{2(-1) - 1 } = \frac{1 + 2 }{-2-1} = \frac{3}{-3} = -1 $
artinya nilai $ \displaystyle \lim_{x \to -1 } \frac{x^2 + 2}{2x - 1 } = -1 $
Sifat-sifat Limit Fungsi
Berikut sifat-sifat limit fungsi :
i). $ \displaystyle \lim_{x \to a } k = k \, $ dengan $ k \, $ adalah konstanta.
ii). $ \displaystyle \lim_{x \to a } k f(x) = k \displaystyle \lim_{x \to a } f(x) $
iii). $ \displaystyle \lim_{x \to a } [f(x) \pm g(x) ] = \displaystyle \lim_{x \to a } f(x) \pm \displaystyle \lim_{x \to a } g(x) $
iv). $ \displaystyle \lim_{x \to a } [f(x). g(x)] = \left( \displaystyle \lim_{x \to a } f(x) \right) \left( \displaystyle \lim_{x \to a } g(x) \right) $
v). $ \displaystyle \lim_{x \to a } \frac{f(x)}{g(x)} = \frac{ \displaystyle \lim_{x \to a } f(x) }{\displaystyle \lim_{x \to a } g(x) } $
vi). $ \displaystyle \lim_{x \to a } [f(x)]^n = \left[ \displaystyle \lim_{x \to a } f(x) \right]^n $
vii). $ \displaystyle \lim_{x \to a } \sqrt[n]{f(x)} = \sqrt[n]{\displaystyle \lim_{x \to a } f(x) } $
i). $ \displaystyle \lim_{x \to a } k = k \, $ dengan $ k \, $ adalah konstanta.
ii). $ \displaystyle \lim_{x \to a } k f(x) = k \displaystyle \lim_{x \to a } f(x) $
iii). $ \displaystyle \lim_{x \to a } [f(x) \pm g(x) ] = \displaystyle \lim_{x \to a } f(x) \pm \displaystyle \lim_{x \to a } g(x) $
iv). $ \displaystyle \lim_{x \to a } [f(x). g(x)] = \left( \displaystyle \lim_{x \to a } f(x) \right) \left( \displaystyle \lim_{x \to a } g(x) \right) $
v). $ \displaystyle \lim_{x \to a } \frac{f(x)}{g(x)} = \frac{ \displaystyle \lim_{x \to a } f(x) }{\displaystyle \lim_{x \to a } g(x) } $
vi). $ \displaystyle \lim_{x \to a } [f(x)]^n = \left[ \displaystyle \lim_{x \to a } f(x) \right]^n $
vii). $ \displaystyle \lim_{x \to a } \sqrt[n]{f(x)} = \sqrt[n]{\displaystyle \lim_{x \to a } f(x) } $
2). Tentukan nilai limit fungsi berikut dengan menggunakan sifat-sifat yang ada,
a). $ \displaystyle \lim_{x \to 2 } 5 $
b). $ \displaystyle \lim_{x \to 3 } 2x^3 $
c). $ \displaystyle \lim_{x \to 1 } x^2 + x $
d). $ \displaystyle \lim_{x \to -1 } x^2 - 3x $
e). $ \displaystyle \lim_{x \to -2 } x^3.x^2 $
f). $ \displaystyle \lim_{x \to 3 } \frac{x^2 - 1}{x + 1} $
g). $ \displaystyle \lim_{x \to 2 } (2x^2 + 3)^9 $
h). $ \displaystyle \lim_{x \to 3 } \sqrt[3]{ x^2 - 1 } $
Penyelesaian :
a). $ \displaystyle \lim_{x \to 2 } 5 = 5 $
b). $ \displaystyle \lim_{x \to 3 } 2x^3 = 2 . \displaystyle \lim_{x \to 3 } x^3 = 2. 3^3 = 2. 37 = 74 $
c). $ \displaystyle \lim_{x \to 1 } x^2 + x = ..... $
$ \begin{align} \displaystyle \lim_{x \to 1 } x^2 + x & = \displaystyle \lim_{x \to 1 } x^2 + \displaystyle \lim_{x \to 1 } x \\ & = 1^2 + 1 \\ & = 1 + 1 = 2 \end{align} $
d). $ \displaystyle \lim_{x \to -1 } x^2 - 3x = ..... $
$ \begin{align} \displaystyle \lim_{x \to -1 } x^2 - 3x & = \displaystyle \lim_{x \to -1 } x^2 - \displaystyle \lim_{x \to -1 } 3x \\ & = \displaystyle \lim_{x \to -1 } x^2 - 3.\displaystyle \lim_{x \to -1 } x \\ & = (-1)^2 - 3.(-1) \\ & = 1 + 3 = 4 \end{align} $
e). $ \displaystyle \lim_{x \to -2 } x^3.x^2 = ..... $
$ \begin{align} \displaystyle \lim_{x \to -2 } x^3.x^2 & = \displaystyle \lim_{x \to -2 } x^3 . \displaystyle \lim_{x \to -2 } x^2 \\ & = (-2)^3 . (-2)^2 \\ & = -8 . 4 = -32 \end{align} $
f). $ \displaystyle \lim_{x \to 3 } \frac{x^2 - 1}{x + 1} = ..... $
$ \begin{align} \displaystyle \lim_{x \to 3 } \frac{x^2 - 1}{x + 1} & = \frac{ \displaystyle \lim_{x \to 3 } x^2 - 1}{ \displaystyle \lim_{x \to 3 } x + 1} \\ & = \frac{ \displaystyle \lim_{x \to 3 } x^2 - \displaystyle \lim_{x \to 3 } 1}{ \displaystyle \lim_{x \to 3 } x + \displaystyle \lim_{x \to 3 } 1} \\ & = \frac{ 3^2 - 1 }{ 3 + 1 } \\ & = \frac{ 8 }{ 4 } = 2 \end{align} $
g). $ \displaystyle \lim_{x \to 2 } (2x^2 + 3)^9 = ..... $
$ \begin{align} \displaystyle \lim_{x \to 2 } (2x^2 + 3)^9 & = \left( \displaystyle \lim_{x \to 2 } 2x^2 + 3 \right)^9 \\ & = \left( \displaystyle \lim_{x \to 2 } 2x^2 + \displaystyle \lim_{x \to 2 } 3 \right)^9 \\ & = \left( 2. \displaystyle \lim_{x \to 2 } x^2 + \displaystyle \lim_{x \to 2 } 3 \right)^9 \\ & = \left( 2. 2^2 + 3 \right)^9 \\ & = \left( 8 + 3 \right)^9 \\ & = \left( 11 \right)^9 \end{align} $
h). $ \displaystyle \lim_{x \to 3 } \sqrt[3]{ x^2 - 1 } = ..... $
$ \begin{align} \displaystyle \lim_{x \to 3 } \sqrt[3]{ x^2 - 1 } & = \sqrt[3]{ \displaystyle \lim_{x \to 3 } x^2 - 1 } \\ & = \sqrt[3]{ \displaystyle \lim_{x \to 3 } x^2 - \displaystyle \lim_{x \to 3 } 1 } \\ & = \sqrt[3]{ 3^2 - 1 } \\ & = \sqrt[3]{ 8 } = 2 \end{align} $
Catatan : Untuk menyelesaikan limit, bisa langsung substitusi saja tanpa harus dipecah menggunakan sifat-sifat yang ada karena hasilnya juga sama.
Contoh :
$ \begin{align} \displaystyle \lim_{x \to 3 } \sqrt[3]{ x^2 - 1 } = \sqrt[3]{ 3^2 - 1 } = \sqrt[3]{ 8 } = 2 . \end{align} $
