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 $1+\cos t$:

$\dfrac{t\sin t}{1-\cos t}\cdot\dfrac{1+\cos t}{1+\cos t}=\dfrac{t\sin t(1+\cos t)}{\sin^2t}=\dfrac{t(1+\cos t)}{\sin t}$

Now,

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

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

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