Evaluate and simplify: $\lim_{x\to 0}\frac{\sin 2x}{3x}$

Question

Evaluate and simplify: $$\lim_{x\to 0}\frac{\sin 2x}{3x}$$

So, by direct substitution, it is indeterminate. I got the derivative of this functio,n and now I'm going around in circles with the trig identities.

Is there a more efficient way of solving this?

Answer 1

Use L'Hôpital's rule

Differentiate both sides

$$\lim_{x\to 0} \frac {2\cos {2x}}{3}$$

Put $x=0$

$\cos {(2*0)}=1$

$$\lim_{x\to 0} \frac {2\cos {2x}}{3}= \frac {2}{3}$$

Answer 2

$$\lim_{x\to0}\frac{\sin 2x}{3x}=\lim_{x\to0}\frac{2}{3}\cdot\frac{\sin 2x}{2x}=\frac{2}{3}\cdot\lim_{x\to0}\frac{\sin 2x}{2x}=\frac{2}{3}$$

Answer 3

Hint: Note that the function $$\frac{\sin y}{y}$$ has a limiting value of $1$ as $y$ vanishes.

Answer 4

Apply L'Hôpital's rule (https://en.wikipedia.org/wiki/L%27Hôpital%27s_rule).

We have $lim_{x\rightarrow 0}\frac{sin2x}{3x} =lim_{x\rightarrow 0}\frac{2cos2x}{3}=\frac{2}{3}.$

One more approach that you may apply using the sine series($Sin(x)=x-\frac{x^3}{3}+\ldots$) and then apply the limit $x\rightarrow 0$, you will get the same result.

Answer 5

Use L'Hospital's Rule:

$$\lim_{x \to 0}\frac{\sin(2x)}{3x}$$ $$\lim_{x \to 0}\frac{2\cos(2x)}{3}$$ $$\lim_{x \to 0}\frac{2*1}{3}$$ $$\frac{2}{3}$$

Answer 6

Another way is to use the identity $$ \sin 2x=2\sin x\cos x $$ whence $$ \lim_{x\to 0}\frac{\sin 2x}{3x}=\lim_{x\to 0}\frac{2\sin x\cos x}{3x}=\frac{2}{3} \left(\lim_{x\to0}\frac{\sin x}{x}\right)\left(\lim_{x\to 0}\cos x\right)=\frac{2}{3}\times 1\times 1=\frac{2}{3}$$ where we used the limit laws in conjunction with the well-known fact that $$ \lim_{x\to0}\frac{\sin x}{x}=1 $$ as well as the fact that cosine is a continuous function.