Question

what is the limit of sin ax/bx

Answer 1

lim x---> 1....................

Answer 2

lim_(x to 0) sin(ax)/sin(bx)
is in 0/0 indeterminate form so we can use l'Hopital's rule = lim_(x to 0) (a cos (ax))/(b cos(bx))
we can lift out the contant term and note that the limit of the quotient is the quotient of the limits where the limits are known =a/b (lim_(x to 0) cos (ax))/(lim_(x to 0) cos(bx))
=a/b lim_(x to 0) (1)/(1) = a/b
Another approach is to head for known limit lim_(z to 0) (sin z)/z = 1
and to proceed as follows#lim_(x to 0) sin(ax)/sin(bx)
=lim_(x to 0) sin(ax)/ x x / (sin(bx))
=lim_(x to 0) (sin(ax)/ x)/( (sin(bx)) /x)
=a/b lim_(x to 0) (sin(ax)/ (ax))/( (sin(bx)) /(bx) )
=a/b (lim_(x to 0) sin(ax)/ (ax))/(lim_(x to 0) (sin(bx)) /(bx) )
with subs#u = ax, v = wz
=a/b (lim_(u to 0) sin(u)/ (u))/(lim_(v to 0) (sin(v)) /(v) )
=a/b *1/1