Find the limit


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


A little prep work gives us


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


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


Combining all this, we have



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


  • $\begingroup$ 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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