# Trig Limit

## Homework Statement

I tried to solve this limit, but failed in the process 'cause the answer's 1 but I get 0. I'd just like for you to check out my steps and tell me what I've done wrong, not post a different solution. Thanks in advance.

$$\mathop {\lim }\limits_{x \to 0} \frac{{2\sin \left( x \right) - \sin \left( {2x} \right)}}{{{x^3}}}$$

## Homework Equations

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( x \right)}}{x} = 1$$

## The Attempt at a Solution

$$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \left( x \right) - \sin \left( {2x} \right)}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \left( x \right)}}{{{x^3}}} - \frac{{\sin \left( {2x} \right)}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \left( 2 \right)\left( {\frac{{\sin \left( x \right)}}{x}} \right)\left( {\frac{1}{{{x^2}}}} \right) - \left( {\frac{{\sin \left( {2x} \right)}}{{2x}}} \right)\left( {\frac{2}{{{x^2}}}} \right) \\ = \mathop {\lim }\limits_{x \to 0} \left( 2 \right)\left( 1 \right)\left( {\frac{1}{{{x^2}}}} \right) - \left( 1 \right)\left( {\frac{2}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{2}{{{x^2}}} - \frac{2}{{{x^2}}} = 0 \\ \end{array}$$

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## The Attempt at a Solution

$$\begin{array}{l} \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \left( x \right) - \sin \left( {2x} \right)}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin \left( x \right)}}{{{x^3}}} - \frac{{\sin \left( {2x} \right)}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \left( 2 \right)\left( {\frac{{\sin \left( x \right)}}{x}} \right)\left( {\frac{1}{{{x^2}}}} \right) - \left( {\frac{{\sin \left( {2x} \right)}}{{2x}}} \right)\left( {\frac{2}{{{x^2}}}} \right)$$
Your mistake is right here. You can't selectively apply limits to some terms but not to others. In this case, you took the limit of $\frac{\sin{x}}{x}$, but not of $\frac{1}{x^2}$ or $\frac{2}{x^2}$.

Have you seen L'Hôpital's rule yet? It seems like the easiest way to solve it to me.

Last edited:
lurflurf
Homework Helper
Why would we want to use L'Hôpital's rule? In most calculus books this would be like a chapter 3 question, and L'Hôpital's rule would be in like chapter 10.

$$\frac{{2\sin x -\sin 2 x }{x^3}}=\left( \frac{{\sin \frac{{x}{2}} }{\frac{{x}{2}}\right) }^3\cos \frac{{x}{2}}$$

Last edited:
Thanks a lot spamiam and lurflurf!

dextercioby
$$\frac{2\sin x - \sin 2x}{x^3} = 2\frac{\sin x}{x} \frac{1-\cos x}{x^2} = 2\frac{\sin x}{x} 2\frac{\sin^2 \frac{x}{2}}{x^2} =\frac{\sin x}{x}\frac{\sin^2 \frac{x}{2}}{\left(\frac{x}{2}\right)^2}$$