Solving Trigonometric Limits: \[\lim_{x\rightarrow 1},\lim_{x\rightarrow -1}\]

[Yankel]

Hello all

I am struggling with these two limits:

\[\lim_{x\rightarrow 1}\frac{sin(x^{2}-1)}{x-1}\]

\[\lim_{x\rightarrow -1}\frac{sin(x^{2}-1)}{x-1}\]

I know that

\[\lim_{x\rightarrow 0}\frac{sin(x)}{x}=1\]

but can't see how it helps me here. I tried multiplying by x+1 both the nominator and the denominator and got

\[\lim_{x\rightarrow 1}\frac{sin(x^{2}-1)}{x^{2}-1}\]

(of course there is a x+1 up there, but that's the easy part)

which is similar to what I know, but x->1 and not x->0, so it doesn't fit.

Can you assist please?

 

[mathbalarka]

Great idea there. You're right that $\lim_{x \to 1} \dfrac{\sin(x^2 - 1)}{(x - 1)} = \lim_{x \to 1} \dfrac{\sin(x^2 - 1)}{(x^2 - 1)} \cdot (x + 1) = 2 \lim_{x \to 1} \dfrac{\sin(x^2 - 1)}{x^2 - 1}$.

To proceed, you have to change the variable in the limit. Note that as $x\to 1$ you have $x^2 - 1 \to 0$, so to evaluate $\lim_{x \to 1} \dfrac{\sin(x^2 - 1)}{x^2 - 1}$, plug in $t = x^2 - 1$, and note that the limit becomes $t \to 0$. Now apply what you know about such limits.

 

[Greg]

You've got the right approach. Provided

$$\lim_{x\to a}f(x)=0$$

then

$$\lim_{x\to a}\frac{\sin(f(x))}{f(x)}=1$$

 

[Yankel]

Nice idea, didn't think of it