$\displaystyle \text{For}\displaystyle \lim_{x \to \infty} \frac{\sin{x}}{x}=\lim_{x \to \infty} \frac{1}{x}\times \lim_{x \to \infty} \sin{x}=0 \times \text{Unknown} $ Is the answer zero?
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Originally Posted by crossbone $\displaystyle \text{For}\displaystyle \lim_{x \to \infty} \frac{\sin{x}}{x}=\lim_{x \to \infty} \frac{1}{x}\times \lim_{x \to \infty} \sin{x}=0 \times \text{Unknown} $ Is the answer zero? $\displaystyle -1\ \le\ sinx\ \le\ 1$ $\displaystyle \displaystyle\lim_{x \to \infty}\ \left(-\frac{1}{x}\right)\ \le\ \lim_{x \to \infty}\frac{sin\;x}{x}\ \le\ \lim_{x \to \infty}\left(\frac{1}{x}\right)$
Thanks! just curious, what's the answer to $\displaystyle \displaystyle \lim_{x \to \infty} sin\;x \ $? Undefined?
Originally Posted by crossbone Thanks! just curious, what's the answer to $\displaystyle \displaystyle \lim_{x \to \infty} sin\;x \ $? Undefined? Yes that limit is not defined.
Thanks! btw, what are indeterminate forms as far as limits is concern? $\displaystyle \frac{0}{0}, \frac{\infty}{\infty}, \frac{0}{\infty}, \frac{\infty}{0}, \infty \times 0$?
Originally Posted by crossbone Thanks! just curious, what's the answer to $\displaystyle \displaystyle \lim_{x \to \infty} sin\;x \ $? Undefined? Can you apply the squeeze theorem to $\displaystyle \displaystyle\lim_{x \to \infty}\frac{sin\;x}{x}\;\;?$ Edit: sorry, the absence of a denominator bypassed my vision! Thanks Plato (next post).
Last edited by Archie Meade; Jan 3rd 2011 at 01:19 PM.
Originally Posted by Archie Meade Can you apply the squeeze theorem to $\displaystyle \displaystyle\lim_{x \to \infty}\sin(x)?$ Not to that limit. It bounces from -1 to 1.
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