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I have this in my lecture:

How did $$\lim_{x\rightarrow \infty} x^3 \left(\tan{\frac{1}{x}}\right)\left(\sin{\frac{3}{x^2}}\right)$$ become $$\lim_{x\rightarrow \infty}3\left(\frac{\tan{\frac{1}{x}}}{\frac{1}{x}}\right)\left(\frac{\sin{\frac{3}{x^2}}}{\frac{3}{x^2}}\right)$$

Note the $x^3$ and the $3$ and denominator of the 2nd equation

UPDATE

The question is how come the $x^3$ became $3\left(\frac{1}{\frac{1}{x}}\right)\left(\frac{1}{\frac{3}{x^2}}\right)$

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What's the question? You're just telling us that you have something in your lecture. So? –  Arturo Magidin Jan 29 '12 at 5:34

1 Answer 1

up vote 5 down vote accepted

It is likely a typo. It should probably read:

$$ 3 \cdot \frac{\tan(\frac{1}{x})}{\frac{1}{x}} \cdot \frac{\sin(\frac{3}{x^2})}{\frac{3}{x^2}} $$ which by multiplying the denominator is $$ \frac{3}{\frac{1}{x} \cdot \frac{3}{x^2}} \cdot \tan\left(\frac{1}{x} \right) \sin\left(\frac{3}{x^2}\right) = x^3 \tan\left(\frac{1}{x} \right) \sin\left(\frac{3}{x^2}\right). $$ More importantly, do you see where these fractions come from? That is, do you wee why we would want to divide $\tan(1/x)$ by $1/x$ and $\sin(3/x^2)$ by $3/x^2$?

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Thanks. Also updated the question with the typo. –  Jiew Meng Jan 29 '12 at 8:12

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