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$$ \lim_{x\to 0}\frac{\sin x \ - \ \sin x \cos x}{x^2\cos x}$$

I graphed it and got zero, and I also tried to solve it algebraically by first factoring out $\dfrac{\sin x}{x} $so that it equals one and you're left with $\dfrac{1-\cos x}{x\cos x}$. I then separated this into two fractions, $\dfrac{1}{x\cos x}-\dfrac{\cos x}{x\cos x}$ canceled the $\cos x $ to get $\dfrac{1}{x \cos x}- \dfrac 1x$, then factored out $\dfrac1x$ to get $\lim \dfrac1x \times \lim \dfrac 1{\cos x} - 1$, which is zero.

So it basically came down to $\lim \dfrac{\sin x}{x}\times \lim \dfrac 1x \times \lim 0$ I thought that even though you get 1, doesn't exist, and zero, respectively, you could just assume the limit was zero since anything times zero is zero. Additionally, I thought this answer was right because when I graphed the function, the limit was indeed zero.

any help would be greatly appreciated! thanks!

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  • $\begingroup$ It's sinx-cosxsinx. sorry, clicked approve edits but didn't realize the problem changed. $\endgroup$ – Melissa Jan 30 '15 at 4:48
  • $\begingroup$ Like that now? (after a refreshing editing break...) There is also a known limit for $ \ \lim_{x \rightarrow 0} \ \frac{1 \ - \ \cos x}{x} \ $ . $\endgroup$ – colormegone Jan 30 '15 at 4:49
  • $\begingroup$ yes, thank you very much! $\endgroup$ – Melissa Jan 30 '15 at 4:50
  • $\begingroup$ You cannot "split" a limit (except if all the limits you write exist)! For example, $\lim_{x\to 0} x/x = 1$ but $1/x$ has no limit in $0$. $\endgroup$ – anderstood Jan 30 '15 at 4:52
  • $\begingroup$ @anderstood It's OK for the problem as has been resolved (there was a bit of "dueling editors" going on here for a little while...). $\endgroup$ – colormegone Jan 30 '15 at 4:54
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Better try splitting it into finite products! $$\lim_{x\to 0}\frac{\sin x-\cos x\sin x}{x^2\cos x}=\lim_{x\to 0}\underbrace{\frac{\sin x}{x}}_{1}\cdot\underbrace{\frac{1-\cos x}{x}}_{0}\cdot\underbrace{\frac1{\cos x}}_{1}=1\cdot0\cdot1=0$$

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  • $\begingroup$ how do you know the limit of 1-cosx/x is zero? $\endgroup$ – Melissa Jan 30 '15 at 4:53
  • $\begingroup$ @Melissa By knowing that the limit of $\frac{1-\cos(x)}{x^2/2}$ equals $1$, which follows from the limit of $\frac{\sin(x)}{x}$ being $1$, by using that $\cos(0)=1$, and adding the two cosines. $\endgroup$ – Pp.. Jan 30 '15 at 4:57
  • $\begingroup$ @Melissa see these links: 1, 2, 3, 4 $\endgroup$ – RE60K Jan 30 '15 at 4:59
  • $\begingroup$ ADG and @Pp.. thank you guys so much for your help! :) It's very much appreciated! $\endgroup$ – Melissa Jan 30 '15 at 5:08
  • $\begingroup$ @Melissa: If this answer helped you, accept it! That is the way to show appreciation to the best answer (and this is the only answer right now). When you get enough reputation you can vote up all answers that help. $\endgroup$ – Rory Daulton Jan 30 '15 at 6:11

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