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Lim x → 0 sin3x)/5x^3 -4

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\bf \lim\limits_(x\to 0)\ \cfrac{sin^3(x)}{5x^3}-4 \\ \quad \\\\ \cfrac{sin^3(x)}{5x^3}-4\implies \cfrac{[sin(x)]^3}{5x^3}-4 \\ \quad \\ \cfrac{[sin(x)]^3}{x^3}\cdot \cfrac{1}{5}-4 \\ \quad \\ thus \\ \quad \\ \lim\limits_(x\to 0)\cfrac{[sin(x)]^3}{x^3}\cdot \lim\limits_(x\to 0)\cfrac{1}{5}-4\qquad \boxed{recall \qquad \lim\limits_(x\to 0) \cfrac{sin(x)}{x}\implies 1} \\ \quad \\ 1\cdot \cfrac{1}{5}-4
User Andy McCluggage
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