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Find the limit $$\lim_{x\to 1}\frac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)}.$$ I'm a little rusty with limits, can somebody please give me some pointers on how to solve this one? Also, l'Hôpital's rule isn't allowed in case you were thinking of using it. Thanks in advance.

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This looks so horrible that I wouldn't even try it without L'Hospital: why would anyone forbid its use? – DonAntonio Oct 15 '12 at 4:36
Change all the functions to sines and cosines; use what you know about $\lim_{a\to0}\sin a/a$; do some algebraic manipulations like factoring. – Gerry Myerson Oct 15 '12 at 4:37
@GerryMyerson I was thinking about doing something like that but I wasn't sure how to proceed. – Chance Oct 15 '12 at 4:39
Try expanding $\sin(a-b)$ and $\cos(a-b)$, you will get some constants and maybe cancellations, then use Gerry's advice on the limit – Alex Oct 15 '12 at 4:42
@Alex, I think expanding the trig functions makes things worse, not better. Chance, Marvis has done more-or-less what I had in mind. – Gerry Myerson Oct 15 '12 at 4:49
up vote 3 down vote accepted

$$\dfrac{\sin(3x-3)}{\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)}{x^2(x-1)^2}$$ Hence, $$\dfrac{(x^2-1)\sin(3x-3)}{\cos(x^3-1)\tan^2(x^2-x)} = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x-1)(x^2-1)}{x^2(x-1)^2 \cos(x^3-1)}\\ = \dfrac{\sin(3x-3)}{3x-3} \times \left(\dfrac{x^2-x}{\tan(x^2-x)} \right)^2 \times \dfrac{3(x+1)}{x^2 \cos(x^3-1)}$$ Now the first and second term on the right has limit $1$ as $x \to 1$. The last term limit can be obtained by plugging $x=1$, to give the limit as $$1 \times 1 \times \dfrac{3 \times (1+1)}{1^2 \times \cos(0)} = 6$$

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I follow what you did, but isn't the sin(x)/x=1 rule only valid if x approaches 0, and this example approaches 1. – Chance Oct 15 '12 at 5:49
@Chance You have $$\lim_{x \to 1} \dfrac{\sin(3x-3)}{3x-3} = \lim_{y \to 0} \dfrac{\sin(y)}y$$ where $y = 3x-3$. – user17762 Oct 15 '12 at 5:53
Yes I forgot that you could do that, it's been a long day. Thank you. – Chance Oct 15 '12 at 5:59

The solution below is the same as the solution by Marvis. Note that as $x\to 1$, $\cos(x^2-1)\to 1$, so we can forget about it. Also, $x^2-1=(x+1)(x-1)$, and the $x+1$ part approaches $2$, so we can sort of forget about it, multiplying by $2$ at the end. And $\tan^2(x^2-x)=\dfrac{\sin^2 (x^2-x)}{\cos^2(x^2-x)}$. The $\cos^2(x^2-x)$ part harmlessly approaches $1$.

So we are interested only in $$\lim_{x\to 1}\frac{(x-1)\sin(3x-3)}{\sin^2(x^2-x)}.$$ I feel better already. Don't really like all those $x-1$, so let $u=x-1$. Then $x^2-x=u(u+1)$ and we are down to $$\lim_{u\to 0}\frac{u\sin 3u}{\sin^2(u(u+1))}.\tag{$1$}$$ Multiply top and bottom by $3(u(u+1))^2$. After minor cancellaton, we want $$\lim_{u\to 0} \frac{3}{(u+1)^2}\cdot\frac{\sin 3u}{3u}\cdot \frac{(u(u+1))^2}{\sin^2(u(u+1))}.$$ We are down to three terms, the first of which has limit $3$, and the last two of which are variants of the standard $\lim_{t\to 0}\dfrac{\sin t}{t}$. So the limit is $3$. Finally, we remember about the factor $2$ that we jettisoned earlier.

Remark: After we reach Expression $1$, the multiplying top and bottom by $3(u(u+1))^2$ is just to please the grader. Near $0$, $\sin t$ and $t$ are first-order indistinguishable, so $\sin 3u\approx 3u$ and $\sin^2 (u(u+1))\approx u^2(u+1)^2$. Indeed, $\tan t$ and $t$ are also first-order indistinguishable, and one can read off the limit without any calculation.

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