Pertanyaan

Tentukan nilai limit fungsi x → π lim sin ( 2 x ) 1 − tan ( x ) − 1 + tan ( x ) .

Tentukan nilai limit fungsi $x \to π lim \frac{1 - tan ( x ) - 1 + tan ( x )}{sin ( 2 x )} .$

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

x → π lim sin ( 2 x ) 1 − tan ( x ) − 1 + tan ( x ) = 0.

x \to π lim \frac{1 - tan ( x ) - 1 + tan ( x )}{sin ( 2 x )} = 0.

Pembahasan

Untuk menjawab soal tersebut, nilai x = π bisa disubstitusi ke dalam fungsi. Substitusi langsung: lim x → π s i n ( 2 x ) 1 − t a n ( x ) − 1 + t a n ( x ) = = s i n ( 2 π ) 1 − t a n ( π ) − 1 + t a n ( π ) 0 0 ( tak tentu ) Jika menghasilkan bentuk tak tentu maka boleh mengubah bentuk fungsi trigonometri: = = = = = s i n ( 2 x ) 1 − t a n ( x ) − 1 + t a n ( x ) × 1 − t a n ( x ) + 1 + t a n ( x ) 1 − t a n ( x ) + 1 + t a n ( x ) s i n ( 2 x ) ( 1 − t a n ( x ) + 1 + t a n ( x ) ) 1 − t a n 2 ( x ) s i n ( 2 x ) ( 1 − t a n ( x ) + 1 + t a n ( x ) ) 1 − t a n 2 ( x ) × c o s 2 ( x ) c o s 2 ( x ) s i n ( 2 x ) c o s 2 ( x ) ( 1 − t a n ( x ) + 1 + t a n ( x ) ) c o s 2 ( x ) − s i n 2 ( x ) s i n ( 2 x ) c o s 2 ( x ) ( 1 − t a n ( x ) + 1 + t a n ( x ) ) c o s ( 2 x ) c o s 2 ( x ) ( 1 − t a n ( x ) + 1 + t a n ( x ) ) t a n ( 2 x ) Sehingga, Dengan demikian, x → π lim sin ( 2 x ) 1 − tan ( x ) − 1 + tan ( x ) = 0.

Untuk menjawab soal tersebut, nilai $x = π$ bisa disubstitusi ke dalam fungsi.

Substitusi langsung:

$lim_{x \to π} \frac{1 - t a n ( x ) - 1 + t a n ( x )}{s i n ( 2 x )} = = \frac{1 - t a n ( π ) - 1 + t a n ( π )}{s i n ( 2 π )} \frac{0}{0} (tak tentu)$

Jika menghasilkan bentuk tak tentu maka boleh mengubah bentuk fungsi trigonometri:

$= = = = = \frac{1 - t a n ( x ) - 1 + t a n ( x )}{s i n ( 2 x )} \times \frac{1 - t a n ( x ) + 1 + t a n ( x )}{1 - t a n ( x ) + 1 + t a n ( x )} \frac{1 - t a n ^{2} ( x )}{s i n ( 2 x ) ( 1 - t a n ( x ) + 1 + t a n ( x ) )} \frac{1 - t a n ^{2} ( x )}{s i n ( 2 x ) ( 1 - t a n ( x ) + 1 + t a n ( x ) )} \times \frac{c o s ^{2} ( x )}{c o s ^{2} ( x )} \frac{c o s ^{2} ( x ) - s i n ^{2} ( x )}{s i n ( 2 x ) c o s ^{2} ( x ) ( 1 - t a n ( x ) + 1 + t a n ( x ) )} \frac{c o s ( 2 x )}{s i n ( 2 x ) c o s ^{2} ( x ) ( 1 - t a n ( x ) + 1 + t a n ( x ) )} \frac{t a n ( 2 x )}{c o s ^{2} ( x ) ( 1 - t a n ( x ) + 1 + t a n ( x ) )}$

Sehingga,

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow straight pi of fraction numerator square root of 1 minus tan open parentheses x close parentheses end root minus square root of 1 plus tan open parentheses x close parentheses end root over denominator sin open parentheses 2 x close parentheses end fraction end cell equals cell limit as x rightwards arrow straight pi of fraction numerator size 12px tan begin mathsize 12px style left parenthesis 2 x right parenthesis end style over denominator size 12px cos to the power of size 12px 2 begin mathsize 12px style left parenthesis x right parenthesis end style begin mathsize 12px style left parenthesis square root of 1 minus tan open parentheses x close parentheses end root plus square root of 1 plus tan open parentheses x close parentheses end root right parenthesis end style end fraction end cell row blank equals cell 0 over 1 end cell row blank equals 0 end table$

Dengan demikian, $x \to π lim \frac{1 - tan ( x ) - 1 + tan ( x )}{sin ( 2 x )} = 0.$