How do I take the limit of this trigonometric function

Question

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?

Answer 1

You can just note that $$ \lim_{a\to 1} \frac{\sin(a^4-1)}{a^3-1} = \lim_{a \to 1}\frac{\sin(a^4-1)}{a^4-1} \cdot \frac{a^4-1}{a^3-1}=\lim_{a\to 1} \frac{\sin(a^4-1)}{a^4-1} \cdot \lim_{a \to 1} \frac{a^4-1}{a^3-1}=1 \times \frac 43= \frac 43 $$

Answer 2

Here is a sexy trick:

$$\frac{\sin(a^4-1)}{a^3-1}=\frac{\sin(a^4-1)}{a^4-1}\times\frac{a^4-1}{a^3-1}$$