Tan(4x)/(sin(6x)) as lim -> 0

what are the steps to solving this WITHOUT using L'hopitals rule?

You can use the limits tanx/x = 1 and sinx/x = 1

lim (x-->0) tan(4x)/sin(6x)

= lim (x-->0) (6x/6x)(4x/4x)(tan(4x)/sin(6x))

= lim (x-->0) (4x/6x)[(6x)(tan(4x))/(4x)(sin(6x))]

= lim (x-->0) (4x/6x)(1)(1)

= lim (x-->0) 4/6, x =/= 0

= 2/3

Let 2x = z ; As x --> 0, z --> 0
Numerator = tan (4x) = tan 2z = sin (2z) / cos (2z) = 2 (sin z)*(cos z) / {1 - 2 (sin z)^2}
As z --> 0, sin z --> z and cos z --> 1
Lim z-->0, Numerator = 2 z / (1 - 2 z^2)
Denominator = sin (6x) = sin 3z = 3 sin z - 4 (sin z)^3
Lim z --> 0, Denominator = 3z - 4z^3 = z(3 - 4z^2)
Hence Lim x --> 0, tan(4x)/{sin(6x)} = {2z / (1 - 2 z^2)} * { 1 / z(3 - 4 z^2)}
As z --> 0 neglecting higher powers of z compared to z, we get the limit as
2z / 3z = 2 / 3

tan(4x) = [sin(4x)] / [ cos(4x)]
= [4 sin(x)cos(x)(2 cos^2(x)-1)] / [cos(4x)]
sin(6x) = [2 sin(x)cos(x)(16 cos^4(x) - 16 cos^2(x) + 3)]
tan(4x)/(sin(6x))
= [2 (2 cos^2(x)-1)] / {[cos(4x)] [(16 cos^4(x) - 16 cos^2(x) + 3)]}

lim(x->0) tan(4x)/(sin(6x))
= lim(x->0) [2 (2 cos^2(x)-1)] / {[cos(4x)] [(16 cos^4(x) - 16 cos^2(x) + 3)]}
= [2(2 - 1)] / [ 1 (16 - 16 + 3)] = 2/3

tan(4x)/(sin(6x)) as lim -> 0
= 4/6
= 2/3
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Ideas: tan x ~ sin x ~ x as x -> 0