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Find the limit

$$\lim_{x\to2}\frac{\sin^2(x^2-4)\sec^2(3x-6)}{(x^3-8)\tan(2x-4)}$$

i have been having trouble finding this limit, i have tried using having trig identities and making all terms sin and cos but i cant figure it out, keep in mind i am not allowed to use L'hopital's rule

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A little prep work gives us

$${\sin^2(x^2-4)\sec^2(3x-6)\over(x^3-8)\tan(2x-4)}={\sin^2(x^2-4)\over(x-2)\sin(2x-4)}\cdot{\cos(2x-4)\over(x^2+2x+4)\cos^2(3x-6)}$$

Now the second term causes no trouble as $x\to2$. It gives

$$\lim_{x\to2}{\cos(2x-4)\over(x^2+2x+4)\cos^2(3x-6)}={\cos0\over12\cos^20}={1\over12}$$

If we rewrite the first term as

$$\begin{align} {\sin^2(x^2-4)\over(x-2)\sin(2x-4)}&={2(x+2)^2(x-2)\sin^2(x^2-4)\over2(x+2)^2(x-2)^2\sin(2x-4)}\\ \\ &={(x+2)^2\over2}\left({\sin(x^2-4)\over(x^2-4)}\right)^2{2x-4\over\sin(2x-4)}\\ \end{align}$$

we can now use the limit

$$\lim_{u\to0}{\sin u\over u}=1$$

with $u=x^2-4$ for one piece and $u=2x-4$ for another to get

$$\lim_{x\to2}{\sin^2(x^2-4)\over(x-2)\sin(2x-4)}={4^2\over2}\cdot1^2\cdot{1\over1}=8$$

Combining all this, we have

$$\lim_{x\to2}{\sin^2(x^2-4)\sec^2(3x-6)\over(x^3-8)\tan(2x-4)}=8\cdot{1\over12}={2\over3}$$

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If you know $\tan x$ ~ $\sin x$ ~ $x$ as $x \to 0$, then this question will be simple.

$$\lim_{x\to2}\frac{\sin^2(x^2-4)\sec^2(3x-6)}{(x^3-8)\tan(2x-4)}=\lim_{x\to2}\frac{(x^2-4)^2}{(x^3-8)(2x-4)}=\frac23.$$

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  • $\begingroup$ isnt the limit as $x$ to $0$ $sinx$ = $0$? why is the term inside $sinx$ still there? $\endgroup$ – katie russo Oct 15 '14 at 3:40

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