Pertanyaan

Tentukan nilai limit berikut. x → ∞ lim ⎝ ⎛ 3 − 9 + 2 1 x 7 x ⎠ ⎞

Tentukan nilai limit berikut.

$x \to \infty lim ⎝ ⎛ \frac{7 x}{3 - 9 + \frac{1}{2} x} ⎠ ⎞$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

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Pembahasan

Mencarinilai limit: Jadi, .

Mencari nilai limit:

$limit as x rightwards arrow infinity thin space of open parentheses fraction numerator 7 x over denominator 3 minus square root of 9 plus 1 half x end root end fraction close parentheses equals 7 times limit as x rightwards arrow infinity thin space of open parentheses fraction numerator x over denominator 3 minus square root of 9 plus 1 half x end root end fraction times fraction numerator 3 plus square root of 9 plus 1 half x end root over denominator 3 plus square root of 9 plus 1 half x end root end fraction close parentheses equals 7 times limit as x rightwards arrow infinity thin space of open parentheses fraction numerator x open parentheses 3 plus square root of 9 plus 1 half x end root close parentheses over denominator 9 minus 9 minus 1 half x end fraction close parentheses equals 7 times limit as x rightwards arrow infinity thin space of open parentheses fraction numerator x open parentheses 3 plus square root of 9 plus 1 half x end root close parentheses over denominator negative 1 half x end fraction close parentheses equals 7 times limit as x rightwards arrow infinity thin space of open parentheses negative 2 open parentheses 3 plus square root of 9 plus 1 half x end root close parentheses close parentheses equals equals 7 times limit as x rightwards arrow infinity thin space of open parentheses negative 2 open parentheses 3 plus square root of 1 half open parentheses x plus 18 close parentheses end root close parentheses close parentheses equals 7 times open parentheses negative 2 close parentheses open parentheses 3 plus square root of 1 half open parentheses x plus 18 close parentheses end root close parentheses equals 7 open parentheses negative 2 close parentheses open parentheses 3 plus square root of 1 half open parentheses infinity close parentheses end root close parentheses equals 7 open parentheses negative 2 close parentheses left parenthesis 3 plus infinity right parenthesis equals 7 open parentheses negative 2 right parenthesis times left parenthesis infinity thin space close parentheses equals negative infinity thin space$

Jadi, $limit as x rightwards arrow infinity thin space of open parentheses fraction numerator 7 x over denominator 3 minus square root of 9 plus 1 half x end root end fraction close parentheses equals negative infinity thin space$ .