Question

limit as x approaches 0 when sinx(1-cosx) / x^2

Answer 1

\[\lim_{x\to0}{\sin x(1-\cos x)\over x^2}=\lim_{x\to0}{\sin x\over x}\cdot\lim_{x\to0}{1-\cos x\over x}\]

Answer 2

are you allowed to use l'Hospital's rule?

Answer 3

Yes, but I don't quite understand the Hospital rule.

Answer 4

I know that six/x = 1 & that 1-cosx/x = 0

Answer 5

If you want to answer it that way that is fine, then you get the limit as \[\lim_{x\to0}\frac{\sin x}x\cdot\lim_{x\to0}\frac{1-\cos x}x=1\cdot0=0\]to prove that you can use l'Hospital:
the first limit evaluates to\[\lim_{x\to0}{\sin x\over x}=\frac00\]so that means we can take the derivative of the numerator and denominator and take the limit again
what is the derivative of the numerator?
what is the derivative of the denominator

Answer 6

same thing with the next term:\[\lim_{x\to0}\frac{1-\cos x}x\]what is the derivative of the numerator?
what is the derivative of the denominator?
replace each piece with its derivative and take the limit again...

Answer 7

My teacher prefers it simple, but thank you for your explanation it really helped (:

Answer 8

sure thing!