Question

Limit of f(t) as t approaches 0. f(t) = (t sin(t)) ÷ (1-cos(t))

Answer 1

Recall the Pythagorean identity,

`1-\cos^2t=\sin^2t`

To get this expression in the fraction, multiply the numerator and denominator by
`1+\cos t`:

`(t\sin t)/(1-\cos t)\cdot(1+\cos t)/(1+\cos t)=(t\sin t(1+\cos t))/(\sin^2t)=(t(1+\cos t))/(\sin t)`

Now,

`\displaystyle\lim_(t\to0)(t\sin t)/(1-\cos t)=\lim_(t\to0)\frac t{\sin t}\cdot\lim_(t\to0)(1+\cos t)`

The first limit is well-known and equal to 1, leaving us with

`\displaystyle\lim_(t\to0)(1+\cos t)=1+\cos0=\boxed{2}`