Pertanyaan

x → 0 lim 1 − cos x x x + 4 − 2 = …

$x \to 0 lim \frac{x x + 4 - 2}{1 - cos x} = \dots$

Y. Endah

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Jadi, jawaban yang tepat adalah B.

$begin mathsize 14px style limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator x square root of x plus 4 end root minus 2 over denominator 1 minus cos invisible function application x end fraction cross times fraction numerator 1 plus cos invisible function application x over denominator 1 plus cos invisible function application x end fraction cross times fraction numerator square root of x plus 4 end root plus 2 over denominator square root of x plus 4 end root plus 2 end fraction close parentheses equals limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator open parentheses x square root of x plus 4 end root minus 2 close parentheses open parentheses 1 plus cos invisible function application x close parentheses over denominator 1 minus cos squared invisible function application x end fraction cross times fraction numerator square root of x plus 4 end root plus 2 over denominator square root of x plus 4 end root plus 2 end fraction close parentheses equals limit as x rightwards arrow 0 of invisible function application fraction numerator x open parentheses x plus 4 minus 4 close parentheses open parentheses 1 plus cos invisible function application x close parentheses over denominator sin squared invisible function application x open parentheses square root of x plus 4 end root plus 2 close parentheses end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator x squared open parentheses 1 plus cos invisible function application x close parentheses over denominator sin squared invisible function application x open parentheses square root of x plus 4 end root plus 2 close parentheses end fraction equals limit as x rightwards arrow 0 of invisible function application open parentheses open parentheses fraction numerator x over denominator sin invisible function application x end fraction close parentheses squared. fraction numerator open parentheses 1 plus cos invisible function application x close parentheses over denominator open parentheses square root of x plus 4 end root plus 2 close parentheses end fraction close parentheses equals 1 squared cross times fraction numerator open parentheses 1 plus cos invisible function application 0 close parentheses over denominator open parentheses square root of 0 plus 4 end root plus 2 close parentheses end fraction equals fraction numerator 1 plus 1 over denominator 2 plus 2 end fraction equals 1 half blank end style$

Jadi, jawaban yang tepat adalah B.