Pertanyaan

Nilai dari x → ∞ lim 4 x − 2 1 − cos ( x 4 ) adalah ...

Nilai dari $x \to \infty lim \frac{1 - cos ( \frac{4}{x} )}{4 x ^{- 2}}$ adalah ...

4
2
1
-4
-2

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A. Khairunisa

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa Sehingga Misalkan , maka . Ketika , maka . Sehingga

Perhatikan bahwa

begin mathsize 14px style cos space space straight a equals 1 minus 2 space sin squared 1 half straight a end style

Sehingga

$begin mathsize 14px style limit as straight x rightwards arrow infinity of fraction numerator 1 minus cos space open parentheses begin display style 4 over straight x end style close parentheses over denominator 4 straight x to the power of negative 2 end exponent end fraction equals limit as straight x rightwards arrow infinity of open parentheses open parentheses 1 minus cos space open parentheses 4 over straight x close parentheses close parentheses space. space straight x squared over 4 close parentheses equals limit as straight x rightwards arrow infinity of open parentheses open parentheses 1 minus open parentheses 1 minus 2 space sin squared space 1 half open parentheses 4 over straight x close parentheses close parentheses close parentheses. space straight x squared over 4 close parentheses equals limit as straight x rightwards arrow infinity of open parentheses open parentheses 2 space sin squared space open parentheses 2 over straight x close parentheses close parentheses space. space straight x squared over 4 close parentheses equals limit as straight x rightwards arrow infinity of open parentheses open parentheses 2 space sin squared space 2 space open parentheses 1 over straight x close parentheses close parentheses space. space straight x squared over 4 close parentheses end style$

Misalkan undefined , maka . Ketika , maka .

Sehingga

$begin mathsize 14px style limit as straight x rightwards arrow infinity of open parentheses open parentheses 2 space sin squared space 2 space open parentheses 1 over straight x close parentheses close parentheses space. space straight x squared over 4 close parentheses equals limit as straight y rightwards arrow 0 of open parentheses open parentheses 2 space sin squared space 2 straight y close parentheses space. space 1 fourth open parentheses 1 over straight y close parentheses squared close parentheses equals limit as straight y rightwards arrow 0 of open parentheses open parentheses 2 space sin squared space 2 straight y close parentheses space. space fraction numerator 1 over denominator 4 straight y squared end fraction close parentheses equals limit as straight y rightwards arrow 0 of open parentheses fraction numerator 2 space sin squared space 2 straight y over denominator 4 straight y squared end fraction close parentheses space equals limit as straight y rightwards arrow 0 of open parentheses fraction numerator 2 space sin squared space 2 straight y space. space sin squared space 2 straight y over denominator 2 straight y space. space 2 straight y end fraction close parentheses space equals limit as straight y rightwards arrow 0 of open parentheses fraction numerator 2 space sin squared space 2 straight y space over denominator 2 straight y space end fraction space. space fraction numerator sin space 2 straight y over denominator 2 straight y end fraction space close parentheses space equals limit as straight y rightwards arrow 0 of open parentheses fraction numerator 2 space sin squared space 2 straight y space over denominator 2 straight y space end fraction space. space fraction numerator sin space 2 straight y over denominator 2 straight y end fraction space close parentheses space equals limit as straight y rightwards arrow 0 of open parentheses 2. space fraction numerator sin squared space 2 straight y space over denominator 2 straight y space end fraction space. space fraction numerator sin space 2 straight y over denominator 2 straight y end fraction space close parentheses space equals 2 space. space 1 space. space 1 equals 2 end style$