Pertanyaan

x → 1 lim x 2 + 3 − x − 1 x 2 − 1

$x \to 1 lim \frac{x ^{2} - 1}{x ^{2} + 3 - x - 1}$

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Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

= = = = = = = = = = = = = = lim x → 1 x 2 + 3 − x − 1 x 2 − 1 lim x → 1 x 2 + 3 − ( x − 1 ) 2 x 2 − 1 lim x → 1 x 2 + 3 − x 2 + 2 x + 1 x 2 − 1 lim x → 1 x 2 + 3 − x 2 + 2 x + 1 x 2 − 1 ⋅ x 2 + 3 + x 2 + 2 x + 1 x 2 + 3 + x 2 + 2 x + 1 lim x → 1 x 2 + 3 − ( x 2 + 2 x + 1 ) ( x 2 − 1 ) ( x 2 + 3 + x 2 + 2 x + 1 ) lim x → 1 x 2 + 3 − x 2 − 2 x − 1 ( x 2 − 1 ) ( x 2 + 3 + x 2 + 2 x + 1 ) lim x → 1 2 − 2 x ( x 2 − 1 ) ( x 2 + 3 + x 2 + 2 x + 1 ) lim x → 1 − 2 ( x − 1 ) ( x + 1 ) ( x − 1 ) ( x 2 + 3 + x 2 + 2 x + 1 ) lim x → 1 − 2 ( x + 1 ) ( x 2 + 3 + x 2 + 2 x + 1 ) − 2 ( 1 + 1 ) ( 1 2 + 3 + 1 2 + 2 ⋅ 1 + 1 ) − 2 2 ( 4 + 4 ) − 2 2 ( 2 + 2 ) − 2 2 ( 4 ) − 2 8 − 4 Jadi, jawaban yang benar adalah A.

$= = = = = = = = = = = = = = lim_{x \to 1} \frac{x ^{2} - 1}{x ^{2} + 3 - x - 1} lim_{x \to 1} \frac{x ^{2} - 1}{x ^{2} + 3 - ( x - 1 ) ^{2}} lim_{x \to 1} \frac{x ^{2} - 1}{x ^{2} + 3 - x ^{2} + 2 x + 1} lim_{x \to 1} \frac{x ^{2} - 1}{x ^{2} + 3 - x ^{2} + 2 x + 1} \cdot \frac{x ^{2} + 3 + x ^{2} + 2 x + 1}{x ^{2} + 3 + x ^{2} + 2 x + 1} lim_{x \to 1} \frac{( x ^{2} - 1 ) ( x ^{2} + 3 + x ^{2} + 2 x + 1 )}{x ^{2} + 3 - ( x ^{2} + 2 x + 1 )} lim_{x \to 1} \frac{( x ^{2} - 1 ) ( x ^{2} + 3 + x ^{2} + 2 x + 1 )}{x ^{2} + 3 - x ^{2} - 2 x - 1} lim_{x \to 1} \frac{( x ^{2} - 1 ) ( x ^{2} + 3 + x ^{2} + 2 x + 1 )}{2 - 2 x} lim_{x \to 1} \frac{( x + 1 ) ( x - 1 ) ( x ^{2} + 3 + x ^{2} + 2 x + 1 )}{- 2 ( x - 1 )} lim_{x \to 1} \frac{( x + 1 ) ( x ^{2} + 3 + x ^{2} + 2 x + 1 )}{- 2} \frac{( 1 + 1 ) ( 1 ^{2} + 3 + 1 ^{2} + 2 \cdot 1 + 1 )}{- 2} \frac{2 ( 4 + 4 )}{- 2} \frac{2 ( 2 + 2 )}{- 2} \frac{2 ( 4 )}{- 2} \frac{8}{- 2} - 4$