limitations and infinity

Question

lim x→0  sin(αx)/sin(βx). For any non zero α and β

Answer 1

As x-> 0,the given expression has the form 0/0. It is indeterminate in form. So we can apply L'Hopital's Rule and take the derivative of the top and bottom.

= lim x->0 (α·cos(αx))/(β·cos(βx)) Now cos(0) =1 so as x->0 the previous expression goes to

(α·1 )/ (β·1) = α/β