XI 6.2 : Sifat Limit Fungsi

Sebelumnya kita sudah membahas tentang definisi limit fungsi dan beberapa contoh mengenai sifat limit, kali ini kita hanya akan memperdalam tentang contoh pada masing-masing sifat limit fungsi.

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B. Sifat-Sifat Limit Fungsi

Selanjutnya kita akan membahas sifat-sifat limit fungsi.

Gunanya apa?

Agar kita bisa menghitung nilai sebuah limit tanpa perlu menggunakan bantuan tabel.

• Misalkan $k, c$ adalah kontanta
• $\displaystyle \lim_{x \to c}f(x)$ ada
• $\displaystyle \lim_{x \to c}g(x)$ ada

Maka berlaku :

1. $\displaystyle \lim_{x \to c}k = k$
2. $\displaystyle \lim_{x \to c}x = c$
3. $\displaystyle \lim_{x \to c}[kf(x)] = k[\displaystyle \lim_{x \to c}f(x)]$
4. $\displaystyle \lim_{x \to c}[f(x) \pm g(x)] = \displaystyle \lim_{x \to c}f(x) \pm \displaystyle \lim_{x \to c}g(x)$
5. $\displaystyle \lim_{x \to c}[f(x).g(x)] = \displaystyle \lim_{x \to c}f(x) . \displaystyle \lim_{x \to c}g(x)$
6. $\displaystyle \lim_{x \to c}[\frac{f(x)}{g(x)}] = \frac{\displaystyle \lim_{x \to c}f(x)}{\lim_{x \to c}g(x)}; \displaystyle \lim_{x \to c}g(x) \neq 0$
7. $\displaystyle \lim_{x \to c}[f(x)]^n = [\displaystyle \lim_{x \to c}f(x)]^n$
8. $\displaystyle \lim_{x \to c}\sqrt[n]{f(x)}=\sqrt[n]{\displaystyle \lim_{x \to c}f(x)}$ dengan $\displaystyle \lim_{x \to c}f(x) \geq 0$ jika $n$ genap

$\odot$ Sifat 1 $\odot$

$\displaystyle \lim_{x \to c}k = k$

Tentukan nilai limit berikut :

a. $\displaystyle \lim_{x \to 2}5 = …$
b. $\displaystyle \lim_{x \to 3}-4 = …$
c. $\displaystyle \lim_{x \to 12}8 = …$
d. $\displaystyle \lim_{x \to 4}7 = …$

Jawab :

Berdasarkan sifat 1, maka nilai limit dari sebuah konstanta adalah tetap konstanta itu sendiri.

a. $\displaystyle \lim_{x \to 2}5 = 5 $
b. $\displaystyle \lim_{x \to 3}-4 = -4$
c. $\displaystyle \lim_{x \to 12}8 = 8$
d. $\displaystyle \lim_{x \to 4}7 = 7$

$\odot$ Sifat 2 $\odot$

$\displaystyle \lim_{x \to c}x = c$

Tentukan nilai limit berikut :

a. $\displaystyle \lim_{x \to 6}x = …$
b. $\displaystyle \lim_{x \to -2}x = …$
c. $\displaystyle \lim_{x \to -4}x^2 = …$
d. $\displaystyle \lim_{x \to 3}x^3 = …$

Jawab :

Sesuai sifat $\displaystyle \lim_{x \to c}x = c$, maka kita subtitusi nilai x menjadi,

a. $\displaystyle \lim_{x \to 6}x = 6$
b. $\displaystyle \lim_{x \to -2}x = -2$
c. $\displaystyle \lim_{x \to -4}x^2 = (-4)^2 = 16$
d. $\displaystyle \lim_{x \to 3}x^3 = 3^3 = 27$

$\odot$ Sifat 3 $\odot$

$\displaystyle \lim_{x \to c}[kf(x)] = k[\displaystyle \lim_{x \to c}f(x)]$

Contoh :

Tentukan nilai limit berikut!

a. $\displaystyle \lim_{x \to 3}4(x+2)= …$
b. $\displaystyle \lim_{x \to 7}2(x-5)= …$
c. $\displaystyle \lim_{x \to -2}3(x^2)= …$

Jawab :

a. $\displaystyle \lim_{x \to 3}4(x+2)= …$
$\begin{aligned}\displaystyle \lim_{x \to 3}4(x+2)&= 4 .\displaystyle \lim_{x \to 3}(x+2)\\ &= 4.(3+2) \\ &= 4 .(5) \\ &= 20 \end{aligned}$

b. $\displaystyle \lim_{x \to 7}2(x-5)= …$
$\begin{aligned}\displaystyle \lim_{x \to 7}2(x-5)&= 2 . \displaystyle \lim_{x \to 7}(x-5)\\ &= 2 .(7-5) \\ &= 2 .(2) \\ &= 4 \end{aligned}$

c. $\displaystyle \lim_{x \to -2}3(x^2)= …$
$\begin{aligned}\displaystyle \lim_{x \to -2}3(x^2)&= 3 . \displaystyle \lim_{x \to -2}(x^2)\\ &= 3 .((-2)^2) \\ &= 3 .(4) \\ &= 12 \end{aligned}$

$\odot$ Sifat 4 $\odot$

$\displaystyle \lim_{x \to c}[f(x) \pm g(x)] = \displaystyle \lim_{x \to c}f(x) \pm \displaystyle \lim_{x \to c}g(x)$

Contoh :

$\begin{aligned}\displaystyle \lim_{x \to 3}(x+2) &= \lim_{x \to 3}x + \lim_{x \to 2}2 \\ &= 3 + 2 \\ &= 5 \end{aligned}$

$\begin{aligned}\displaystyle \lim_{x \to -1 } x^2 – 3x & = \displaystyle \lim_{x \to -1 } x^2 – \displaystyle \lim_{x \to -1 } 3x \\ & = \displaystyle \lim_{x \to -1 } x^2 – 3.\displaystyle \lim_{x \to -1 } x \\ & = (-1)^2 – 3.(-1) \\ & = 1 – (-3) \\ & = 1 + 3 \\ &= 4\end{aligned}$

$\begin{aligned}\displaystyle \lim_{x \to 1 } x^2 + x &= \displaystyle \lim_{x \to 1 } x^2 + \displaystyle \lim_{x \to 1 } x \\& = 1^2 + 1 \\ & = 1 + 1 \\ &= 2 \end{aligned}$

$\displaystyle \begin{aligned} \lim_{x \to 3} \left( x^{3} – 2x \right) &= \lim_{x \to 3} x^{3} – \lim_{x \to 3} 2x \\ &= 3^{3} – 2 . \lim_{x \to 3} x \\ &= 27 – 2 . 3 \\ &= 27 – 6 \\ &= 21 \end{aligned}$

$\displaystyle \begin{aligned} \lim_{x \to 3} \left( x^{2} + 2x \right) &= \lim_{x \to 3} x^{2} + \lim_{x \to 3} 2x \\ &= 3^{2} + 2 . \lim_{x \to 3} x \\ &= 9 + 2 . 3 \\ &= 9 + 6 \\ &= 15 \end{aligned}$

$\displaystyle \begin{aligned} \lim_{x \to 3} \left( x^{2} +8x – 6 \right) &= \lim_{x \to 3} x^{2} + \lim_{x \to 3} 8x – \lim_{x \to 3} 6 \\ &= 3^{2} + 8 . \lim_{x \to 3} x – 6 \\ &= 9 + (8 . 3) – 6 \\ &= 9 + 24 – 6\\ &= 27 \end{aligned}$

$\odot$ Sifat 5 $\odot$

$\displaystyle \lim_{x \to c}[f(x).g(x)] = \displaystyle \lim_{x \to c}f(x) . \displaystyle \lim_{x \to c}g(x)$

Contoh :

$\displaystyle \begin{aligned} \lim_{x \to -2} \left( x^{2} . x \right) &= \lim_{x \to -2} x^{2} . \lim_{x \to -2} x \\ &= (-2)^{2} . (-2) \\ &= 4 . (-2) \\ &= -8 \end{aligned}$

$\begin{aligned} \displaystyle \lim_{x \to 2 }\left( x^{3} . x^{2} \right) & = \displaystyle \lim_{x \to 2 } x^3 . \displaystyle \lim_{x \to 2 } x^2 \\ & = (2)^3 . (2)^2 \\ & = 8 . 4 \\ &= 32 \end{aligned}$

$\begin{aligned} \displaystyle \lim_{x \to -2 }(x^3 + 3)(x^2-5x) & = \displaystyle \lim_{x \to -2 } (x^3 +3) . \displaystyle \lim_{x \to -2 } (x^2-5x) \\ & = [\lim_{x \to -2 } x^3 + \lim_{x \to -2 }3)][(\lim_{x \to -2 }x^2 – \lim_{x \to -2 }5x)] \\ &= ((-2)^3 + 3)(2^2 – 5 \lim_{x \to -2 }x) \\ &= ((-8)+3)(4-5(-2)) \\ &= (-5)(4-(-10)) \\& = (-5)(4+10) \\ &= (-5)(14) \\ &= -70 \end{aligned}$

$\odot$ Sifat 6 $\odot$

$\displaystyle \lim_{x \to c}[\frac{f(x)}{g(x)}] = \frac{\displaystyle \lim_{x \to c}f(x)}{\lim_{x \to c}g(x)}; \displaystyle \lim_{x \to c}g(x) \neq 0$

Contoh :

$\begin{aligned}\displaystyle \lim_{x \to 1} \left[ \frac{x^{2} + 3}{x+1} \right] &= \frac{\displaystyle \lim_{x \to 1} (x^{2} + 3)}{\displaystyle \lim_{x \to 1} (x+1)} \\ &= \frac{\displaystyle \lim_{x \to 1} x^{2} + \lim_{x \to 1} 3}{\displaystyle \lim_{x \to 1} x+ \lim_{x \to 1} 1} \\ &= \frac{ 1^{2} + 3}{ 1+ 1} \\ &= \frac{ 1 + 3}{2} \\ &= \frac{4}{2} \\ &= 2 \end{aligned}$

$\begin{aligned}\displaystyle \lim_{x \to 3 } \frac{x^2 – 1}{x + 1} & = \frac{ \displaystyle \lim_{x \to 3 } x^2 – 1}{ \displaystyle \lim_{x \to 3 } x + 1} \\ & = \frac{ \displaystyle \lim_{x \to 3 } x^2 – \displaystyle \lim_{x \to 3 } 1}{ \displaystyle \lim_{x \to 3 } x + \displaystyle \lim_{x \to 3 } 1} \\ & = \frac{ 3^2 – 1 }{ 3 + 1 } \\ & = \frac{ 8 }{ 4 } \\ &= 2\end{aligned}$

$\odot$ Sifat 7 $\odot$

$\displaystyle \lim_{x \to c}[f(x)]^n = [\displaystyle \lim_{x \to c}f(x)]^n$

Contoh :

$\begin{aligned} \displaystyle \lim_{x \to 2 } (2x^2 + 3)^9 & = \left( \displaystyle \lim_{x \to 2 } 2x^2 + 3 \right)^9 \\ & = \left( \displaystyle \lim_{x \to 2 } 2x^2 + \displaystyle \lim_{x \to 2 } 3 \right)^9 \\ & = \left( 2. \displaystyle \lim_{x \to 2 } x^2 + \displaystyle \lim_{x \to 2 } 3 \right)^9 \\ & = \left( 2. 2^2 + 3 \right)^9 \\ & = \left( 8 + 3 \right)^9 \\ & = \left( 11 \right)^9 \end{aligned}$

$\begin{aligned}\displaystyle \lim_{x \to 1} \left[ 3x^{2} – 1 \right]^{5} &= \left[ \lim_{x \to 1} (3x^{2} – 1) \right]^{5} \\&\displaystyle = \left[ \lim_{x \to 1} 3x^{2} – \lim_{x \to 1} 1 \right]^{5} \\ &= \left[ 3(1)^{2} -1 \right]^{5} \\ &= \left[ 3 -1 \right]^{5} \\ &= \left[ 2 \right]^{5} \\ &= 32 \end{aligned}$

$\odot$ Sifat 8 $\odot$

$\displaystyle \lim_{x \to c}\sqrt[n]{f(x)}=\sqrt[n]{\displaystyle \lim_{x \to c}f(x)}$ dengan $\displaystyle \lim_{x \to c}f(x) \geq 0$ jika $n$ genap

Contoh :

$\begin{aligned}\displaystyle \lim_{x \to 3 } \sqrt[3]{ x^2 – 1 } & = \sqrt[3]{ \displaystyle \lim_{x \to 3 } x^2 – 1 } \\ & = \sqrt[3]{ \displaystyle \lim_{x \to 3 } x^2 – \displaystyle \lim_{x \to 3 } 1 } \\ & = \sqrt[3]{ 3^2 – 1 } \\ & = \sqrt[3]{ 8 } \\ &= 2 \end{aligned}$

$\begin{aligned}\displaystyle \lim_{x \to 5} \sqrt[3] {x^{2} + 2} &= \displaystyle \sqrt[3] {\lim_{x \to 5} (x^{2} + 2}) \\ &= \displaystyle \sqrt[3] {\lim_{x \to 5} x^{2} + \lim_{x \to 5} 2} \\ &= \sqrt[3] {5^{2} + 2} \\ &= \sqrt[3] {25 + 2} \\ &= \sqrt[3] {27} \\ &= \sqrt[3] {3^{3}} \\ &= 3^{\frac{3}{3}} \\ &= 3^{1} \\ &= 3 \end{aligned}$

Oke, semoga bermanfaat