<s>[INST] How do I take the limit of this trigonometric function I was wondering how I could take this limit:
$\lim_{a→1}\frac{\sin(a^4 - 1)}{a^3-1}$
My idea was that if I can get the denominator and the inside of the sin to be the same I can use the sandwich theorem or maybe cancel out the a-1 in the denominator somehow.  To go about this I did the following operations:
$$\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a-1)(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1) * (a + 1)(a^2 + 1)}{(a-1)(a+1)(a^2+1)*(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1) * (a + 1)(a^2 + 1)}{(a^4 - 1)*(a^2+a+1)}$$
$$\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a^4 - 1)} *\lim_{a\to 1}\frac{(a^2 + 1)(a + 1)}{(a^2 + a + 1)}$$
At this step I tried to use the sandwich theorem to find $\lim_{a\to 1}\frac{\sin(a^4 - 1)}{(a^4 - 1)}$
However I got $(-\infty,\infty)$ Like so:
$$\lim_{a\to 1} -1 < \sin(a^4-1) < 1$$
$$\lim_{a\to 1} \frac{-1}{a^4 - 1} < \frac{\sin(a^4-1)}{a^4-1} < \frac{1}{a^4 - 1}$$
$$= -\infty < \frac{\sin(a^4-1)}{a^4-1} < \infty$$
So I'm at a loss as to what I can do from here on.  Or is what I've done just completely wrong?
 [/INST] Here is a sexy trick:
$$\frac{\sin(a^4-1)}{a^3-1}=\frac{\sin(a^4-1)}{a^4-1}\times\frac{a^4-1}{a^3-1}$$
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