Evaluate the following limit: the limit as θ approaches 0, of sin(3θ)/(θ^2-2θ). DO NOT USE L'HOSPITAL'S RULE

Question

The limit as θ approaches 0, of sin(3θ)/θ²-2θ. Without using L'Hospital's rule.

After applying the derivative using the quotient rule, I got (3cos(3θ))(θ²-2θ)-(2θ-2)(sin(3θ))/(θ²-2θ)²

Do I distribute the sin and cos after this?

Answer 1

You don't need to take a derivative if you are not allowed to use L'Hospital's rule. Instead, you can use the fact that

lim(x->0) sin(x)/x = 1

sin(3θ)/(θ²-2θ) = sin(3θ) / θ(θ-2) - factored the denominator.

Rewriting it as a product of 2 fractions:

sin(3θ)/θ * 1/(θ-2)

To use the fact above, we need to get 3θ in the denominator, since it should match sin(3θ). Multiplying both the numerator and the denominator by 3, we get

lim (θ->0) 3 sin(3θ)/3θ * 1/(θ-2)

Breaking it up into 2 limits, we get:

3 * lim (θ->0) sin(3θ)/3θ * lim (x->0) 1/(θ-2)

Now, lim (θ->0) sin(3θ)/3θ = 1

The last limit = -1/2, just plug in θ=0.

3*1*(-1/2)= - 3/2

lim(θ->0) sin(3θ)/(θ²-2θ) = - 3/2

Answer 2

Recall: lim_x→0 (sin x / x) = 1

lim_θ→0 [sin(3θ) / (θ² - 2θ)]

= lim_θ→0 [(sin(3θ) / θ)(1 / (θ-2)]

= lim_θ→0 [3{sin(3θ)/(3θ)) (1 / (θ-2)] = (3)(1)(-1/2) = -3/2