Nilai lim_(x→-2)⁡ ((x^2-4) tan⁡(x+2))/sin^2⁡(x+2) =⋯

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Bahas Soal Matematika   »   Limit   ›  

Nilai \( \displaystyle \lim_{x\to -2} \ \frac{(x^2-4) \tan(x+2)}{\sin^2 (x+2)} = \cdots \)

  1. -4
  2. -3
  3. 0
  4. 4

Pembahasan:

\begin{aligned} \lim_{x\to -2} \ \frac{(x^2-4) \tan(x+2)}{\sin^2 (x+2)} &= \lim_{x\to -2} \ \frac{(x-2)(x+2) \tan(x+2)}{\sin^2 (x+2)} \\[8pt] &= \lim_{x\to -2} \ (x-2) \cdot \lim_{x\to -2} \ \frac{(x+2)}{\sin (x+2)} \cdot \lim_{x\to -2} \ \frac{\tan(x+2)}{\sin(x+2)} \\[8pt] &= (-2-2) \cdot 1 \cdot 1 \\[8pt] &= -4 \end{aligned}

Jawaban A.