Limit of a function as it tends to zero

Question

$$\lim_{x \to 0}\frac{3(\tan 2x)^2} {2x^2}$$ What method is best for solving this? The division by zero is causing me issues when trying to apply Sandwich theorem and/or Algebra of limits. (I believe I am not suppose to use L'Hopitals to answer this also)

Answer 1

I usually recommend to use basic results when dealing with simple limits. In this case here is my solution: $$\lim_{x \to 0} \frac{3(\tan 2x)^2}{2x^2} = \lim_{x \to 0} \frac{3 \sin 2x \sin 2x}{2x^2 \cos 2x \cos 2x} = \lim_{x \to 0} \frac{\sin 2x}{2x}\frac{\sin 2x}{2x} \frac{6}{\cos^2 2x} = 6,$$ since $\frac{\sin 2x}{2x} \to 1$ and $\cos^2 2x \to 1$ when $x$ goes to $0$. To be more precise, if you put $y = 2x$, $x \to 0$ implies $y \to 0$, so $$\lim_{x \to 0} \frac{\sin 2x}{2x} = \lim_{y \to 0} \frac{\sin y}{y}$$ and it is well known that this is $1$. A similar trick can be applied to $\cos^2 2x$.

Answer 2

Since $\displaystyle \lim_{x\to 0} \dfrac{\tan x}{x} = \tan'(0) = 1+\tan^2(0) = 1$, we obtain, letting $u:=2x$:

$$\dfrac{3(\tan (2x))^2}{2x^2} = 3\dfrac{\tan^2 u}{u^2/2} = 6\left(\dfrac{\tan u}{u}\right)^2$$

thus the limit is $6$.

Answer 3

Write it in the form $$6\frac{\sin(2x)\sin(2x)}{2x\cdot 2x}\cdot \frac{1}{\cos^2(2x)}$$