Question

can anyone tell me why the limit of (cosΔx-1)/Δx when Δx approaches zero is zero?
Thanks a lot!

Answer 1

The answer is in the following attachment:

Answer 2

Using Taylor series for cos x near 0:
\[\cos x = 1 - x^2/2\ + O(x^4)\] Then it's obvious:
\[\lim_{x \rightarrow 0}[(\cos x - 1)/ x ]=\lim_{x \rightarrow 0}[(1 - x^2/2+O(x^4) - 1)/ x ]=\lim_{x \rightarrow 0}[-x/2+O(x^3)]=0\]

Answer 3

if you get a limit that when you plug in the x--->a some a, and you get 0/0 or infinity/infinity you may apply brother L'Hopitals rule which states.

take the d/dt of the top and bottom seperately ie dont do the quotient rule.

so since the limit x-->0 cos(x)-1/x
d/dx of the top and bottom yields
-sin(x)/1 and evaluating this at x=0 yeilds 0/1 which on a good day is 0.