Nilai lim_(x→0)⁡ x^3/(√(1+sin⁡ x)-√(1+tan⁡ x))=⋯

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Bahas Soal Matematika   »   Limit   ›  

Nilai \( \displaystyle \lim_{x \to 0} \ \frac{x^3}{\sqrt{1+\sin x} - \sqrt{1+\tan x}} = \cdots \)

  1. -4
  2. -2
  3. 0
  4. 2
  5. 4

(SBMPTN 2016)

Pembahasan:

\begin{aligned} \lim_{x \to 0} \ \frac{x^3}{\sqrt{1+\sin x} - \sqrt{1+\tan x}} &= \lim_{x \to 0} \ \frac{x^3}{\sqrt{1+\sin x} - \sqrt{1+\tan x}} \cdot \frac{\sqrt{1+\sin x} + \sqrt{1+\tan x}}{\sqrt{1+\sin x} + \sqrt{1+\tan x}} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{(1+\sin x) - (1+\tan x)} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x - \tan x} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x \ (1-\frac{1}{\cos x})} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x \ (\frac{\cos x - 1}{\cos x})} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \cos x \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x \ (\cos x - 1)} \\[8pt] &= \lim_{x \to 0} \ \frac{x^3 \cos x \ (\sqrt{1+\sin x} + \sqrt{1+\tan x})}{\sin x \ (-2 \sin^2 \frac{1}{2}x)} \\[8pt] &= -\frac{1}{2} \cdot \lim_{x \to 0} \ \frac{x}{\sin x} \cdot \lim_{x \to 0} \ \frac{x^2}{\sin^2 \frac{1}{2}x} \cdot \lim_{x \to 0} \ \cos x \ (\sqrt{1+\sin x} + \sqrt{1+\tan x}) \\[8pt] &= -\frac{1}{2} \cdot 1 \cdot \left( \frac{1}{\frac{1}{2}} \right)^2 \cdot \left(\cos 0 \ (\sqrt{1+\sin 0} + \sqrt{1+\tan 0})\right) \\[8pt] &= -\frac{1}{2} \cdot 1 \cdot (2)^2 \cdot \left(1 \ (\sqrt{1+0} + \sqrt{1+0})\right) \\[8pt] &= -2 \cdot (\sqrt{1} + \sqrt{1}) = -4 \end{aligned}

Jawaban A.