Question #60ff3

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Question #60ff3

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Answer 1 · 2017-09-03T21:41:37

Break it down as follows

Explanation:

$\frac{{sin}^{2} (3 x)}{x^{2} cos x} = \frac{{sin}^{2} (3 x)}{x^{2}} \cdot \frac{1}{cos x}$ $sin^2(3x)/(x^2cosx) = sin^2(3x)/(x^2)*1/cosx$
$= {(\frac{sin (3 x)}{x})}^{2} \cdot \frac{1}{cos x}$ $= (sin(3x)/x)^2*1/cosx$
$= 9 \cdot {(\frac{sin (3 x)}{3 x})}^{2} \cdot \frac{1}{cos x}$ $= 9*(sin(3x)/(3x))^2*1/cosx$

Now, as $x \to 0$ $x rarr 0$ , $3 x \to 0$ $3x rarr 0$ , and therefore
$\frac{sin (3 x)}{3 x} \to 1$ $sin(3x)/(3x) rarr 1$
This means that its square also goes to 1.

Since cosine is continuous at 0, $x \to 0$ $x rarr 0$ , $cos x \to 1$ $cosx rarr 1$

Put this together, and you should arrive at an overall limit of 9.

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Question #60ff3

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