Name:
Mark
2006-09-01 13:53
how do i derive lim as x->0 of
(sin5x)/(3x)?
Name:
Anonymous
2006-09-02 0:38
For 0 <= x < pi/2
sin(5x) <= 5x <= tan(5x)
sin(5x)/sin(5x) <= 5x/sin(5x) <= tan(5x)/sin(5x)
1 <= 5x/sin(5x) <= 1/cos(5x)
Lim x->0 of 1 = 1, lim x->0 of 1/cos(5x) = 1, so by Squeeze theorem, lim x->0+ of 5x/sin(5x) is 1.
So, lim x->0+ of sin(5x)/5x = 1, and sin(5x)/3x = 5/3 * sin(5x)/5x, so lim x->0 of sin(5x)/3x = 5/3 * lim x->0+ of sin(5x)/5x = 5/3 * 1 = 5/3.
For -pi/2 < x <= 0
sin(5x) >= 5x >= tan(5x)
(the rest is similar to the above)
Once you have lim x->0+ sin(5x)/3x = 5/3 and lim x->0- sin(5x)/3x = 5/3, you know lim x->0 sin(5x)/3x = 5/3.