Pertanyaan

x → ∞ lim x ( sec x 1 − 1 ) = ...

$x \to \infty lim x (sec \frac{1}{x} - 1) = ...$

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Habis dalam

Klaim

F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

Jawaban terverifikasi

Pembahasan

Misalkan maka untuk maka

$begin mathsize 12px style limit as x rightwards arrow infinity of x open parentheses s e c space fraction numerator 1 over denominator square root of x end fraction minus 1 close parentheses equals limit as x rightwards arrow infinity of x open parentheses fraction numerator 1 over denominator cos open parentheses begin display style fraction numerator 1 over denominator square root of x end fraction end style close parentheses end fraction minus 1 close parentheses equals limit as x rightwards arrow infinity of x open parentheses fraction numerator 1 minus cos open parentheses fraction numerator 1 over denominator square root of x end fraction close parentheses over denominator cos open parentheses begin display style fraction numerator 1 over denominator square root of x end fraction end style close parentheses end fraction close parentheses end style$

Misalkan $begin mathsize 12px style u equals fraction numerator 1 over denominator square root of x end fraction end style$ maka untuk maka

$begin mathsize 12px style limit as u rightwards arrow 0 of 1 over u squared open parentheses fraction numerator 1 minus cos space u over denominator cos space u end fraction close parentheses equals limit as u rightwards arrow 0 of fraction numerator begin display style 1 end style over denominator begin display style u squared end style end fraction open parentheses fraction numerator begin display style 1 minus cos space u end style over denominator begin display style cos space u end style end fraction close parentheses times fraction numerator 1 plus cos space u over denominator 1 plus cos space u end fraction equals limit as u rightwards arrow 0 of fraction numerator begin display style 1 minus cos squared space u end style over denominator begin display style u squared open parentheses cos space u close parentheses open parentheses 1 plus cos space u close parentheses end style end fraction equals fraction numerator 1 over denominator open parentheses cos begin display style space end style begin display style 0 end style close parentheses begin display style open parentheses 1 plus cos space 0 close parentheses end style end fraction equals fraction numerator 1 over denominator open parentheses 1 close parentheses begin display style open parentheses 1 plus 1 close parentheses end style end fraction equals 1 half end style$