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allsm [11]
3 years ago
13

Lim x → 0 sin3x)/5x^3 -4

Mathematics
1 answer:
JulsSmile [24]3 years ago
8 0
\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
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See attachment for missing details

Required

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