1. ## trig limit(2)

$\displaystyle \lim_{x\to1}\frac{sin(x^2-1)}{x-1}$

I tried making t=x-1

but I was left with sin(t(x+1)) which I didn't know what to do with

2. Originally Posted by akhayoon
$\displaystyle \lim_{x\to1}\frac{sin(x^2-1)}{x-1}$

I tried making t=x-1

but I was left with sin(t(x+1)) which I didn't know what to do with
t = x - 1 we get $\displaystyle \lim_{t\to 0} \frac{\sin t(t+2)}{t}$.

We get, $\displaystyle \lim_{t\to 0}\frac{\sin t^2 \cos 2t + \cos t^2\sin 2t}{t}$

Replace $\displaystyle \cos 2t$ and $\displaystyle \cos t^2$ by 1 because that is their limit and we have:

$\displaystyle \lim_{t\to 0}\frac{\sin t^2}{t} + \frac{\sin 2t}{t} = \lim_{t\to 0}\frac{\sin t^2}{t^2}\cdot t + \frac{\sin 2t}{2t} \cdot 2 = 2$

3. There are many ways to approach these limits.

You could let $\displaystyle t=x^{2}-1$

$\displaystyle x=\sqrt{t+1}$

$\displaystyle \lim_{t\rightarrow{0}}\frac{sin(t)}{\sqrt{t+1}-1}$

Now, multiply by the conjugate:

$\displaystyle \lim_{t\rightarrow{0}}\frac{\sqrt{t+1}sin(t)}{t}+\ lim_{t\rightarrow{0}}\frac{sin(t)}{t}$

We can see the familiar sin(t)/t limit. Which equals 1.

So you have the limit 1+1=2

4. sorry,I don't understand how you got $\displaystyle \sqrt{t+1}-1$

nor do I understand how you manipulated the rest of the equation to make those to limits at the end

5. Here's another attempt.

Originally Posted by akhayoon
$\displaystyle \lim_{x\to1}\frac{sin(x^2-1)}{x-1}$
$\displaystyle \lim_{x \to 1} \frac{{\sin \left( {x^2 - 1} \right)}} {{x - 1}} = \lim_{x \to 1} \frac{{\sin \left( {x^2 - 1} \right)}} {{x^2 - 1}} \cdot (x + 1).$

You can see the limit of the first factor exists and it's equal to 1, just make $\displaystyle u=x^2-1$ to check it.

From here you can easily see that the requested value of the limit is 2.

6. Originally Posted by akhayoon
$\displaystyle \lim_{x\to1}\frac{sin(x^2-1)}{x-1}$

I tried making t=x-1

but I was left with sin(t(x+1)) which I didn't know what to do with
$\displaystyle \lim_{x\to{1}}\frac{\sin(x^2-1)}{x-1}=\lim_{x\to{1}}\frac{\sin(x^2-1)-\sin(1^2-1)}{x-1}=\bigg(\sin(x^2-1)\bigg)'\bigg|_{x=1}=2$

That was a tad simpler I think