# Evaluate the limit $\lim_{x\to 2} \frac {\sin(x^2 -4)}{x^2 - x -2}$

$$\lim_{x\to 2} \frac {\sin(x^2 -4)}{x^2 - x -2}$$

Attempt at solution:

So I know I can rewrite denominator:

$$\frac {\sin(x^2 -4)}{(x-1)(x+2)}$$

So what's next? I feel like I'm supposed to multiply by conjugate of either num or denom.... but by what value...?

Don't tell me I'm simply supposed to plug in $x = 2$

I need to simplify fractions somehow first, how?

• Can you use the L'Hôpital's rule? – user226387 May 5 '15 at 16:17
• Your factorisation is incorrect: it should be $(x+1)(x-2)$ (which makes the problem non-trivial, of course). – Chappers May 5 '15 at 16:17
• The factorization of the bottom is wrong. Find the right factorization and multiply top and bottom by $x+2$. – André Nicolas May 5 '15 at 16:18

$$\frac {\sin(x^2 -4)}{(x-2)(x+1)}\cdot\frac{x+2}{x+2} = \frac{(x+2)\sin(x^2 - 4)}{(x+1)(x^2 - 4)} = \dfrac{x+2}{x+1}\cdot \dfrac{\sin(x^2 - 4)}{x^2 - 4}$$
$$\frac{\sin(x^2-4)}{(x+1)(x-2)}=\frac{\sin(x^2-4)}{x^2-4}\,\frac{x^2-4}{(x+1)(x-2)}$$