Teorema Limit

Teorema A : Teorema Dasar Limit
Jika n bilangan bulat positif, k konstan, f dan g fungsi yang mempunyai limit di c, maka berlaku

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ k = k

2.

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ x = c

3.

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ k f(x) = k

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ f(x)

4.

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ [ f(x) + g(x) ] =

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ f(x) +

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ g(x)

5.

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ [ f(x) − g(x) ] =

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ f(x) −

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ g(x)

6.

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ [ f(x) . g(x) ] =

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ f(x) .

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ g(x)

7.

_{x \to c}^{l i m} \frac{f (x)}{g (x)}

$\mathrm{_{x \to c}^{lim}\frac{f(x)}{g(x)}}$ =

\frac{_{x \to c}^{l i m} f (x)}{_{x \to c}^{l i m} g (x)}

$\mathrm{\frac{_{x \to c}^{lim}\,f(x)}{_{x \to c}^{lim}\,g(x)}}$ , dengan

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ g(x) ≠ 0

8.

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ [ f(x) ]ⁿ =

{[_{x \to c}^{l i m} f (x)]}^{n}

$\mathrm{\left [_{x \to c}^{lim}\,f(x) \right ]^{n}}$

9.

_{x \to c}^{l i m} \sqrt[n]{f (x)}

$\mathrm{_{x \to c}^{lim}\sqrt[\mathrm{n}]{\mathrm{f(x)}}}$ =

\sqrt[n]{_{x \to c}^{l i m} f (x)}

$\sqrt[\mathrm{n}]{\mathrm{_{x \to c}^{lim}\,f(x)}}$
dengan

_{x \to c}^{l i m}

$\mathrm{_{x \to c}^{lim}}$ f(x) > 0 ketika n genap.

Contoh 1
Hitung limit berikut dengan menggunakan teorema dasar limit !
a.

_{x \to 3}^{l i m}

$\mathrm{_{x \to 3}^{lim}}$ (2x + 3)
Jawab :

_{x \to 3}^{l i m}

$\mathrm{_{x \to 3}^{lim}}$ (2x + 3) =

_{x \to 3}^{l i m}

$\mathrm{_{x \to 3}^{lim}}$ 2x +

_{x \to 3}^{l i m}

$\mathrm{_{x \to 3}^{lim}}$ 3 (teorema A.4)

_{x \to 3}^{l i m}

$\mathrm{_{x \to 3}^{lim}}$ (2x + 3) = 2

_{x \to 3}^{l i m}

$\mathrm{_{x \to 3}^{lim}}$ x + 3 (A.3 dan A.1)

_{x \to 3}^{l i m}

$\mathrm{_{x \to 3}^{lim}}$ (2x + 3) = 2 . 3 + 3 (A.2)

_{x \to 3}^{l i m}

$\mathrm{_{x \to 3}^{lim}}$ (2x + 3) = 9

b.

_{x \to 5}^{l i m} \sqrt{x^{2} - 16}

$\mathrm{_{x \to 5}^{lim}\sqrt{x^{2}-16}}$
Jawab :

_{x \to 5}^{l i m} \sqrt{x^{2} - 16}

$\mathrm{_{x \to 5}^{lim}\sqrt{x^{2}-16}}$ =

\sqrt{_{x \to 5}^{l i m} (x^{2} - 16)}

$\mathrm{\sqrt{_{x \to 5}^{lim}\,(x^{2}-16)}}$ (A.9)

_{x \to 5}^{l i m} \sqrt{x^{2} - 16}

$\mathrm{_{x \to 5}^{lim}\sqrt{x^{2}-16}}$ =

\sqrt{_{x \to 5}^{l i m} x^{2} -_{x \to 5}^{l i m} 16}

$\mathrm{\sqrt{_{x \to 5}^{lim}\,x^{2}-\,_{x \to 5}^{lim}\,16}}$ (A.5)

_{x \to 5}^{l i m} \sqrt{x^{2} - 16}

$\mathrm{_{x \to 5}^{lim}\sqrt{x^{2}-16}}$ =

\sqrt{{[_{x \to 5}^{l i m} x]}^{2} - 16}

$\mathrm{\sqrt{\left [_{x \to 5}^{lim}\,x \right ]^{2}-\,16}}$ (A.8 dan A.1)

_{x \to 5}^{l i m} \sqrt{x^{2} - 16}

$\mathrm{_{x \to 5}^{lim}\sqrt{x^{2}-16}}$ =

\sqrt{5^{2} - 16}

$\mathrm{\sqrt{5^{2}-\,16}}$ (A.2)

_{x \to 5}^{l i m} \sqrt{x^{2} - 16}

$\mathrm{_{x \to 5}^{lim}\sqrt{x^{2}-16}}$ = 3

Contoh 2

Jika

_{x \to a}^{l i m}

$\mathrm{_{x \to a}^{lim}}$ f(x) = 3 dan

_{x \to a}^{l i m}

$\mathrm{_{x \to a}^{lim}}$ g(x) = 8, tentukan nilai dari

_{x \to a}^{l i m} \frac{f^{2} (x) - g (x)}{2 f (x) + \sqrt[3]{g (x)}}

$\mathrm{_{x \to a}^{lim}\frac{f^{2}(x)\,-\,g(x)}{2f(x)\,+\,\sqrt[3]{\mathrm{g(x)}}}}$
Jawab :

_{x \to a}^{l i m} \frac{f^{2} (x) - g (x)}{2 f (x) + \sqrt[3]{g (x)}}

$\mathrm{_{x \to a}^{lim}\frac{f^{2}(x)-g(x)}{2f(x)+\sqrt[3]{\mathrm{g(x)}}}}$ =

\frac{{[_{x \to a}^{l i m} f (x)]}^{2} -_{x \to a}^{l i m} g (x)}{2_{x \to a}^{l i m} f (x) + \sqrt[3]{_{x \to a}^{l i m} g (x)}}

$\mathrm{\frac{\left [_{x \to a}^{lim}\,f(x) \right ]^{2}\;-\;_{x \to a}^{lim}\,g(x)}{2\,_{x \to a}^{lim}\,f(x)\;+\;\sqrt[3]{\mathrm{_{x \to a}^{lim}\,g(x)}}}}$

_{x \to a}^{l i m} \frac{f^{2} (x) - g (x)}{2 f (x) + \sqrt[3]{g (x)}}

$\mathrm{_{x \to a}^{lim}\frac{f^{2}(x)-g(x)}{2f(x)+\sqrt[3]{\mathrm{g(x)}}}}$ =

\frac{3^{2} - 8}{2.3 + \sqrt[3]{8}}

$\frac{3^{2}\;-\;8}{2.3\;+\;\sqrt[3]{8}}$

_{x \to a}^{l i m} \frac{f^{2} (x) - g (x)}{2 f (x) + \sqrt[3]{g (x)}}

$\mathrm{_{x \to a}^{lim}\frac{f^{2}(x)-g(x)}{2f(x)+\sqrt[3]{\mathrm{g(x)}}}}$ =

\frac{1}{8}

$\frac{1}{8}$

Teorema B : Teorema Substitusi
Jika f adalah fungsi polinom atau fungsi rasional dan f terdefinisi di c, maka

lim x \to c f (x) = f (c)

$\mathrm{\lim_{x\rightarrow c}f(x)=f(c)}$
Contoh 3
Jika f(x) = x³ − 3x, tentukan

_{x \to 2}^{l i m}

$\mathrm{_{x \to 2}^{lim}}$ f(x)
Jawab :
Perhatikan bahwa f(x) adalah fungsi polinom dan kita tahu bahwa fungsi polinom terdefinisi untuk setiap x bilangan real. Jadi, f(x) terdefinisi di 2, akibatnya

_{x \to 2}^{l i m}

$\mathrm{_{x \to 2}^{lim}}$ (x³ − 3x) = 2³ − 3.2 = 2

Contoh 4
Jika g(x) =

\frac{x^{2} + x - 6}{x - 2}

$\mathrm{\frac{x^{2}+x-6}{x-2}}$ , tentukan

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ g(x)
Jawab :
Perhatikan bahwa g(x) adalah fungsi rasional dan kita tahu bahwa fungsi rasional terdefinisi untuk setiap x bilangan real kecuali nilai-nilai x yang menyebabkan penyebutnya bernilai nol, yaitu x = 2. Jadi, g(x) terdefinisi di 1, akibatnya

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$

\frac{x^{2} + x - 6}{x - 2}

$\mathrm{\frac{x^{2}\,+\,x\,-\,6}{x\,-\,2}}$ =

\frac{1^{2} + 1 - 6}{1 - 2}

$\mathrm{\frac{1^{2}\,+\,1\,-\,6}{1\,-\,2}}$ = 4

Contoh 5
Diketahui f(x) =

⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2 x j i k a x < 1 x^{2} j i k a 1 \leq x < 2 x + 2 j i k a x \geq 2 \end{matrix}

$\left\{\begin{matrix} \mathrm{2x\,\,\,\,\,\,\,\,\,\,jika\,x<1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \mathrm{x^{2}\,\,\,\,\,\,\,\,\,\,jika\,\,1\leq x< 2}\,\,\,\,\, \\ \mathrm{x+2\,\,jika\,x\geq 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{matrix}\right.$
Tentukan limit berikut jika ada !
a.

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ f(x)
b.

_{x \to 2}^{l i m}

$\mathrm{_{x \to 2}^{lim}}$ f(x)

Jawab :
a. Untuk x < 1, f(x) = 2x, sehingga

_{x \to 1^{-}}^{l i m}

$\mathrm{_{x \to 1^{-}}^{lim}}$ f(x) =

_{x \to 1^{-}}^{l i m}

$\mathrm{_{x \to 1^{-}}^{lim}}$ 2x = 2 . 1 = 2

Untuk x > 1, f(x) = x², sehingga

_{x \to 1^{+}}^{l i m}

$\mathrm{_{x \to 1^{+}}^{lim}}$ f(x) =

_{x \to 1^{+}}^{l i m}

$\mathrm{_{x \to 1^{+}}^{lim}}$ x² = 1² = 1

Limit kiri ≠ limit kanan, akibatnya

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ f(x) tidak ada

b. Untuk x < 2, f(x) = x², sehingga

_{x \to 2^{-}}^{l i m}

$\mathrm{_{x \to 2^{-}}^{lim}}$ f(x) =

_{x \to 2^{-}}^{l i m}

$\mathrm{_{x \to 2^{-}}^{lim}}$ x² = 2² = 4

Untuk x > 2, f(x) = x + 2, sehingga

_{x \to 2^{+}}^{l i m}

$\mathrm{_{x \to 2^{+}}^{lim}}$ f(x) =

_{x \to 2^{+}}^{l i m}

$\mathrm{_{x \to 2^{+}}^{lim}}$ (x + 2) = 2 + 2 = 4

Limit kiri = limit kanan = 4, akibatnya

_{x \to 2}^{l i m}

$\mathrm{_{x \to 2}^{lim}}$ f(x) = 4

Teorema C
Jika f(x) = g(x) ketika x ≠ c, maka

lim x \to c f (x) = lim x \to c g (x)

$\mathrm{\lim_{x\rightarrow c}f(x)=\lim_{x\rightarrow c}g(x)}$ asalkan limitnya ada.

Contoh 6
Hitung

_{x \to 2}^{l i m} \frac{(x - 2) (x + 3)}{x - 2}

$\mathrm{_{x \to 2}^{lim}\,\frac{(x\,-\,2)(x\,+\,3)}{x\,-\,2}}$
Jawab :
Karena

\frac{(x - 2) (x + 3)}{x - 2}

$\mathrm{\frac{(x\,-\,2)(x\,+\,3)}{x\,-\,2}}$ = x + 3, ketika x ≠ 2, akibatnya

_{x \to 2}^{l i m} \frac{(x - 2) (x + 3)}{x - 2}

$\mathrm{_{x \to 2}^{lim}\,\frac{(x\,-\,2)(x\,+\,3)}{x\,-\,2}}$ =

_{x \to 2}^{l i m}

$\mathrm{_{x \to 2}^{lim}}$ (x + 3)

_{x \to 2}^{l i m} \frac{(x - 2) (x + 3)}{x - 2}

$\mathrm{_{x \to 2}^{lim}\,\frac{(x\,-\,2)(x\,+\,3)}{x\,-\,2}}$ = 2 + 3

_{x \to 2}^{l i m} \frac{(x - 2) (x + 3)}{x - 2}

$\mathrm{_{x \to 2}^{lim}\,\frac{(x\,-\,2)(x\,+\,3)}{x\,-\,2}}$ = 5

Contoh 7
Diketahui f(x) =

\frac{| x - 1 |}{x - 1}

$\mathrm{\frac{\left | x-1 \right |}{x-1}}$ . Hitung

_{x \to 1}^{l i m} f (x)

$\mathrm{_{x \to 1}^{lim}\,f(x)}$ jika ada !

Jawab :
Berdasarkan definisi nilai mutlak :
|x − 1| = x − 1 jika x ≥ 1
|x − 1| = −(x − 1) jika x < 1

Untuk x < 1, f(x) =

\frac{- (x - 1)}{x - 1}

$\mathrm{\frac{-(x-1) }{x-1}}$ = -1, sehingga

_{x \to 1^{-}}^{l i m}

$\mathrm{_{x \to 1^{-}}^{lim}}$ f(x) =

_{x \to 1^{-}}^{l i m} (- 1)

$\mathrm{_{x \to 1^{-}}^{lim}\,(-1)}$ = -1

Untuk x > 1, f(x) =

\frac{x - 1}{x - 1}

$\mathrm{\frac{x-1 }{x-1}}$ = 1, sehingga

_{x \to 1^{+}}^{l i m}

$\mathrm{_{x \to 1^{+}}^{lim}}$ f(x) =

_{x \to 1^{+}}^{l i m} (1)

$\mathrm{_{x \to 1^{+}}^{lim}\,(1)}$ = 1

Limit kiri ≠ limit kanan, akibatnya

_{x \to 1}^{l i m} \frac{| x - 1 |}{x - 1}

$\mathrm{_{x \to 1}^{lim}\frac{\left | x-1 \right |}{x-1}}$ tidak ada

Teorema D : Teorema Apit
Misalkan f, g dan h adalah fungsi-fungsi yang memenuhi f(x) ≤ g(x) ≤ h(x) untuk setiap x di dekat a, kecuali mungkin di a.

lim x \to c f (x) = lim x \to c h (x) = L \Rightarrow lim x \to c g (x) = L

$\mathrm{\lim_{x\rightarrow c}f(x)=\lim_{x\rightarrow c}h(x)=L\;\;\Rightarrow\;\; \lim_{x\rightarrow c}g(x)=L}$

Contoh 8
Jika untuk setiap x berlaku 2x ≤ f(x) ≤ x² + 1, hitunglah

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ f(x)
Jawab :
2x ≤ f(x) ≤ x² + 1

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ 2x = 2 . 1 = 2

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ (x² + 1) = 1² + 1 = 2

Karena

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ 2x =

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ (x² + 1) = 2, berdasarkan teorema apit kita simpulkan

_{x \to 1}^{l i m}

$\mathrm{_{x \to 1}^{lim}}$ f(x) = 2

Contoh 9
Gunakan teorema apit untuk menunjukkan bahwa

_{x \to 0}^{l i m}

$\mathrm{_{x \to 0}^{lim}}$ x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ = 0
Jawab :
Nilai sin θ selalu berada pada interval -1 dan 1 berapapun θ. Secara matematis kita tulis
-1 ≤ sin θ ≤ 1

Jika θ =

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ , maka -1 ≤ sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ ≤ 1

Jika setiap ruas dikalikan dengan x akan diperoleh 2 kasus berikut :
Kasus 1 : Untuk x > 0 diperoleh
-x ≤ x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ ≤ x
Karena

_{x \to 0^{+}}^{l i m}

$\mathrm{_{x \to 0^{+}}^{lim}}$ -x =

_{x \to 0^{+}}^{l i m}

$\mathrm{_{x \to 0^{+}}^{lim}}$ x = 0, sesuai teorema apit kita simpulkan

_{x \to 0^{+}}^{l i m}

$\mathrm{_{x \to 0^{+}}^{lim}}$ x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ = 0 ..........(1)

Kasus 2 : Untuk x < 0 diperoleh
-x ≥ x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ ≥ x atau dapat pula ditulis
x ≤ x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ ≤ -x
Karena

_{x \to 0^{-}}^{l i m}

$\mathrm{_{x \to 0^{-}}^{lim}}$ x =

_{x \to 0^{-}}^{l i m}

$\mathrm{_{x \to 0^{-}}^{lim}}$ -x = 0, sesuai teorema apit kita simpulkan

_{x \to 0^{-}}^{l i m}

$\mathrm{_{x \to 0^{-}}^{lim}}$ x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ = 0 ...........(2)

Dari persamaan (1) dan (2), dapat kita lihat bahwa limit kiri dan limit kanan x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ untuk x menuju 0 nilainya sama, yaitu 0. Akibatnya

_{x \to 0}^{l i m}

$\mathrm{_{x \to 0}^{lim}}$ x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ = 0

Dari grafiknya jelas terlihat, kurva y = x sin

(\frac{1}{x})

$\mathrm{\left (\frac{1}{x} \right )}$ diapit oleh garis y = -x dan y = x dan ketika x mendekati nol, nilai limitnya dipaksa untuk sama dengan nilai limit kedua garis tersebut.

Teorema Limit

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