## gowtham3214 Group Title Hi. In lecture 3, in the derivation of the limit of (1-cos Θ)/Θ as Θ-->0 . I am able to understand tht the 'gap' 1-cosΘ nears 0 as Θ tends to zero. But then, doesnt it mean we re actually ending up with 0/0 rather than 0. can anyone clarify on this ? 3 years ago 3 years ago

1. alistair

This is explained in Session 8 Course Notes Clip 3, Questions and Answers. Basically, as Θ tends to zero, the numerator (1-cos Θ) shrinks faster than the denominator Θ, giving a limit of 0. (Think of Θ as the length of the arc here).

2. gowtham3214

@alistair : thank you :) that helped :) clarified :)

3. luckey

do one thing multiply and divide the whole limit $(1-\cos \theta)\div \theta$ by an additional $\theta$ then the read the limit as $\lim_{\theta \rightarrow 0} ((1-\cos \theta)\div \theta ^{2}) \theta$ solve the limit as $\lim_{\theta \rightarrow 0} ((1-\cos \theta)\div \theta ^{2}$ it'll come put to be 1/2 then the limit is left as $\lim_{\theta \rightarrow 0} \theta$*1/2 this is obviously equals to '0'