Is my sin(2x) over x^3 limit solution actually correct?

Question

I had an exam question with a sin(2x) limit divided by x^3:

\lim_{x \to 0} \frac{\sin(2x) - 2\sin x}{x^3}

I wrote:

Using LHopitals rule:

\lim_{x \to 0} \frac{2\cos(2x) - 2\cos x}{3x^2}

\lim_{x \to 0} \frac{-4\sin(2x) + 2\sin x}{6x}

\lim_{x \to 0} \frac{-8\cos(2x) + 2\cos x}{6}

= \frac{-8(1) + 2(1)}{6} = \frac{-6}{6} = -1

My teacher marked it wrong and said answer is 0. But I applied LHopital 3 times since it was 0/0 each time. Where did I go wrong?

I think it might be because $\sin x \approx x - \frac{x^3}{6}$ and Taylor series might give a different answer. Please help me understand!

David Kim · Accepted Answer

For a trigonometric limit, the first check is whether direct substitution gives a defined value. If substitution gives 0 / 0, then you need to transform the expression before evaluating it. The most common tools are: - standard limits such as lim ⁡ x → 0 sin ⁡ x x = 1, - identities such as 1 − cos ⁡ x = 2 sin ⁡ 2 ( x / 2 ), - factoring, - or L'Hopital's rule if it is allowed in your course. The mistake many students make is canceling terms before they are factors. For example, you cannot cancel sin ⁡ x from: sin ⁡ x + x . You can only cancel common factors from products or fractions. So to check whether your teacher was right, rewrite every line and ask: did I use an identity, factor, or valid limit theorem? If one step only "looks reasonable" but is not algebraically justified, that is probably where the error is.

Is my sin(2x) over x^3 limit solution actually correct?

1 Answer

Is my sin(2x) over x^3 limit solution actually correct?

1 Answer