I haven’t done this type of math problem since the Stone Age

The question was how to do lim(x->0) 3sin4x/sin3x without using L’Hopital’s rule.

The only thing I could think of offhand was to expand out sin4x in terms of sin3x, cos3x, sinx, and cosx, and then work from there. It worked, but was a little messy.

Note: lim is as x->0

lim (3sin4x/sin3x) = 3 lim [ (sin3xcosx + cos3xsinx)/sin3x]

=3 lim [cos x + (cos3xsinx)/sin3x]

= 3 lim[cos x] + 3 lim[ (cos3xsinx)/sin3x]

= 3(1) + 3 lim (cos3xsinx/sin3x)

(a) = 3[1 + lim (cos3xsinx/sin3x ) ]

We need to find (b) lim (cos3xsinx/sin3x ) ]

sin3x = sin2xcosx + cos2xsinx
= (2sinxcosx)cosx + (cos^2x-sin^2x)sinx
= 2sinxcos^2x + cos^2xsinx – sin^3x
= 3sinxcos^3x – sin^3x
=3sinx(cos^3x-sin^2x)

Note: cos^2x is (cosx)^2

(b) is lim [ (cos3xsinx)/ 3sinx(cos^3x-sin^2x) ]

= lim [ cos3x / 3(cos^3x-sin^2x) ]

= 1/[ 3(1-0)]

= 1/3

(b) lim (cos3xsinx/sin3x ) ] = 1/3

Substituting back into (a), the answer is 3(1 + 1/3) =4

This is the correct answer. I wonder if there isn’t an easier way which I might have seen if I were still in college…

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