Question

limit x->0 of x sins/cosx - 1

Answer 1

I'd multiply by the conjugate first. \(\cos x +1 \)

Answer 2

Quite a few ways to go about this.

Answer 3

Well, if you use l'Hospital's rule it might be a lot easier.

Answer 4

I tend to assume they haven't learned it.

Answer 5

Is it 1-coax/x = 0 ? I just don't know how to apply it to this question

Answer 6

Nope, first multiply top and bottom by \(\cos x+1\)

Answer 7

Hmm, I was mistaken. I was thinking \(\sin x\approx x\) near 0, but that leads to indeterminate form again @wio

Answer 8

Though you could use \(\sin x\approx x\) and have
\[\frac{\sin^2x}{1-\cos x}=\frac{1-\cos^2x}{1-\cos x}\]
Then factor, etc

Answer 9

\[
\begin{split}
\lim_{x\to 0} \frac{x\sin x}{\cos x - 1}
&=\lim_{x\to 0} \frac{x\sin x(\cos x +1)}{\cos^2 x - 1}\\
&=\lim_{x\to 0} \frac{x\sin x(\cos x +1)}{-\sin^2 x}\\
&=\lim_{x\to 0} -\frac{x(\cos x +1)}{\sin x}\\
&=\lim_{x\to 0} -\frac{x}{\sin x}(\cos x+1)\\
&=\lim_{x\to 0} -\left(\frac{\sin x}{x}\right)^{-1} (\cos x+1)
\end{split}
\]

Answer 10

Wtf fractions aren't working anymore for me.

Answer 11

\[
\frac{1}{2}
\]

Answer 12

\[\frac{ a }{ s }\]