# can you multiply a limit that doesn't exist times a limit that is equal to zero?

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

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!

• It's sinx-cosxsinx. sorry, clicked approve edits but didn't realize the problem changed. – Melissa Jan 30 '15 at 4:48
• Like that now? (after a refreshing editing break...) There is also a known limit for $\ \lim_{x \rightarrow 0} \ \frac{1 \ - \ \cos x}{x} \$ . – colormegone Jan 30 '15 at 4:49
• yes, thank you very much! – Melissa Jan 30 '15 at 4:50
• 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 $0$. – anderstood Jan 30 '15 at 4:52
• @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...). – colormegone Jan 30 '15 at 4:54

Better try splitting it into finite products! $$\lim_{x\to 0}\frac{\sin x-\cos x\sin x}{x^2\cos x}=\lim_{x\to 0}\underbrace{\frac{\sin x}{x}}_{1}\cdot\underbrace{\frac{1-\cos x}{x}}_{0}\cdot\underbrace{\frac1{\cos x}}_{1}=1\cdot0\cdot1=0$$
• @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. – Pp.. Jan 30 '15 at 4:57