Find the Limits (L'Hospital's rule is not permitted)

Question

a) lim x-->0 sin(6x) / tan(2x)

b) lim x-->0 sin(x-2) / x^2 - 2x

c) lim x-->0 tan(x) + 3x / sin(2x)

Answer 1

Use the property that lim t-->0 sin(t) / t = 1.

a) sin(6x) = 6x * [ sin(6x) / 6x ] and 1 / tan(2x) = cos(2x) / sin(2x) = cos(2x) * [ 2x / sin(2x) ] / 2x.

So lim x-->0 sin(6x) / tan(2x) = [ 6x / 2x ] * [ sin(6x) / 6x ] * cos(2x) * [ 2x / sin(2x) ] = 3 * 1 * 1 * 1 = 3.

b) sin(x -2) / (x²-2x) = sin(x-2) / x(x-2) = [ sin(x-2) / (x-2) \ / x.

Then lim x-->0 sin(x -2) / (x²-2x) = lim x-->0 [sin(x -2) / (x-2)] / x = lim x-->0 1/x.

But lim x-->0 1/x does not exist because the limits from left and right are not equal.

c) lim x-->0 { tan(x) + 3x / sin(2x) } = lim x-->0 { tan(x) + 3x * [ 2x / sin(2x) ] / 2x} = 0 + 3/2 = 3/2.