lim x -->0 sinax/sinbx?

Answer 1

Have you learned L'Hopital's rule yet?

If not, write the fraction as

[sin(ax)/ax] (ax) / {[sin(bx)/bx](bx)}

which equals

(a/b) [sin(ax)/ax] /[sin(bx)/bx]

and use the fact that the limit of the quotient is the limit of the numerator, divided by the limit of the denominator (provided both limits exist, and provided that the limit of the denominator is not zero).

Answer 2

sin(ax) = ax - (ax)^3/3! + (ax)^5/5! -.... +...

=x*(a - a^3*x^2/3! + a^5*x^4/5! -.... +...)

similarly sin(bx) = x*(b - b^3*x^2/3! + b^5*x^4/5! -.... +...)

so that sin(ax) / sin(bx) =

(a - a^3*x^2/3! + a^5*x^4/5! -....)/(b - b^3*x^2/3! + b^5*x^4/5! -....)

now it should be clear that as
x => 0, sin(ax) / sin(bx) => a/b

hth