Recall the Pythagorean identity,
[tex]1-\cos^2t=\sin^2t[/tex]
To get this expression in the fraction, multiply the numerator and denominator by [tex]1+\cos t[/tex]:
[tex]\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}[/tex]
Now,
[tex]\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)[/tex]
The first limit is well-known and equal to 1, leaving us with
[tex]\displaystyle\lim_{t\to0}(1+\cos t)=1+\cos0=\boxed{2}[/tex]
