Problem with evalulating a limit

[John45]

\(\displaystyle \text{I'm having difficulty on where to start with this equation: }\:\lim_{x \to \ 0}\frac{cos^2 (x) - 1}{x}\)
\(\displaystyle \text{I know that }\:\lim_{x \to \ 0}\frac{sin(x)}{x} \;=1\)

\(\displaystyle \text{and }\:\lim_{x \to \ 0}\frac{cos(x)-1}{x} \;=0\)

\(\displaystyle \text{I'm wanting to turn it into}\frac{(1-sin^2(x))-1}{x}\)

\(\displaystyle \text{But then I get stuck on what that did to help and where to go next}\)

 

[daon]

\(\displaystyle \:\lim_{x \to 0}\frac{cos^2(x)-1}{x} = \:\lim_{x \to 0} -\frac{sin^2x}{x} = -\left( \:\lim_{x \to 0} sinx \right )\left ( \:\lim_{x \to 0} \frac{sinx}{x} \right )\)


But why can we do the last separation? That should be justified in your answer.

 

[John45]

So was my process correct and the one's eliminated each other leaving -sin^2 (x) / x ? Or was it taking the derivative of cos?

 

[BigGlenntheHeavy]

\(\displaystyle Another \ way, \ to \ wit:\)

\(\displaystyle \lim_{x\to0}\frac{cos^2(x)-1}{x} \ gives \ the \ indeterminate \ form \ \frac{0}{0}, \ hence \ the \ Marqui \ to\)

\(\displaystyle the \ rescue.\)

\(\displaystyle Therefore, \ \lim_{x\to0}\frac{cos^2(x)-1}{x} \ = \ \lim_{x\to0}[-2sin(x)cos(x)] \ = \ 0.\)

 

[daon]

So was my process correct and the one's eliminated each other leaving -sin^2 (x) / x ? Or was it taking the derivative of cos?

You may think of it that way, yes.

You may also use the second limit you posted by factoring the numerator as (cosx+1)(cosx-1).