Limits of Trig Functions. Calc 1 help

lim (x->0) (sin 2x) / (sin 3x)

Any takers?

lim (x->0) (sin 2x) / (sin 3x)
Divide Numerator and denominator by x..
lim (x->0) (sin kx) /x =k
Use the above relation:
The ans: 2/3

The above rule by matsci is quite a wonderful one that always seems to work with limits that involve sine and cosine. Another rule that you might want to look into when doing problems like these especially when the limit is approaching 0 is L'Hopitals rule.

What L'Hopitals rule tells is that if lim x->a f(x)/g(x) = f(a)/g(a) = 0/0 or ∞/∞, we may evaluate the limit by doing lim x->a f(x)/g(x) = lim x->a f '(x)/g '(x).

So in this problem:

lim x->0 (sin 2x)/(sin 3x) = 0/0 So we must resort to L'Hopitals rule.

lim x->0 2*cos(2x)/3*cos(3x) = 2/3

We use following limit: lim[x→0] sin(x)/x = 1

Then we show: lim[x→0] sin(ax)/x = a

lim[x→0] sin(ax)/x
= lim[x→0] a sin(ax) / (ax)
= a * lim[x→0] sin(ax)/(ax) ..... let u = ax, then as x→0, u→0
= a * lim[u→0] sin(u)/u
= a * 1
= a

lim[x→0] {sin(2x)/sin(3x)}
= lim[x→0] {[sin(2x)/x] / [sin(3x)/x]}
= lim[x→0] {sin(2x)/x} / lim[x→0] {sin(3x)/x}
= 2/3