calculus - Why does the following limit give two answers?

I want to calculate

$$\lim_{t \to 0} \frac{t^2}{\sin^2(t)}$$

and I proceed as follows

$$\stackrel{H}{=} \lim_{t \to 0} \frac{2t}{2\sin(t)\cos(t)} \implies \lim_{t \to 0} \frac{2t}{\sin(2t)}$$

and when evaluated gives

$$\stackrel{H}{=} \lim_{t \to 0} \frac{2}{2\cos^2(t)-2\sin^2(t)} =1$$

But evaluating the other equivalent term gives

$$\stackrel{H}{=} \lim_{t \to 0} \frac{2}{2\sin(2t)\cos(2t)}$$

and that does not exist as the left hand and right hand limits are not equal.

So, what do you think?