why is this true? $\lim_{x\to 0}\frac{\sin^2(x)}{x^2} = \lim_{x\to 0}\frac{\sin(x)}{x}$ Let's say I get to this 
$$\lim_{x\to 0}\frac{\sin^2(x)}{x^2} = \lim_{x\to 0}\frac{\sin(x)}{x} = 1$$
How do I justify that first reduction?  Do you intuit that, or is there some algebraic cancel/reduction involved?
Again, why is this true? 
$$\lim_{x\to 0}\frac{\sin^2(x)}{x^2} = \lim_{x\to 0}\frac{\sin(x)}{x}$$
 A: Hint If $\lim_{x\to a} f(x)=l$ Then $\lim_{x\to a} [f(x)]^2=l^2$
more generally If $\lim_{x\to a} f(x)=l$ and $\lim_{x\to a} g(x)=m$ Then
$\lim_{x\to a} [f(x)\times g(x)]=[\lim_{x\to a} f(x)]\times [\lim_{x\to a} g(x)]$
A: It's only true by sheer luck. We know by the product rule of limits $$lim_{x\rightarrow 0} \frac {\text{sin}^2(x)} {x^2} = \left(\lim_{x \rightarrow 0}\frac {\text{sin}(x)} x \right)^2 = 1^2 = 1$$
If the inside limit had been anything other than $0$ or $1$ - i.e., the solutions to $x^2=x$ - that would have been wrong.
A: Since function $$f(y) = y^2 \ , \quad y \in \mathbb{R}$$ is continuous and you know that $\frac{\sin x}{x} \to 1$ as $x \to 0\ ,$ it follows that $f(\frac{\sin}{x}) \to f(1)$ as $x \to 0\ .$ In other words, $$\lim_{x\to 0} \frac{\sin^2x}{x^2} = 1^2 \ . $$
A: The magic words are "L'Hopital" and "double angle formula". The first tells you that 
$$\lim \frac{\sin^2 x}{x^2} = \lim \frac{2 \sin x \cos x}{2 x}.$$
The second tells you that $2\sin x \cos x = \sin 2 x.$
A: Just use some algebra. Because 
$$
\lim_{x \to 0} \frac{\sin x}{x}=1
$$
we have 
$$
1\cdot 1 \cdot \cdots 1=\lim_{x \to 0} \frac{\sin x}{x}\lim_{x \to 0} \frac{\sin x}{x}\cdots\lim_{x \to 0} \frac{\sin x}{x}=\lim_{x \to 0} \frac{\sin^n x}{x^n}=1
$$
where the product is taken $n$ times.
