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I'm wondering how to solve the following limit: $$\lim\limits_{x\to1}\frac{\tan(x^2-1)}{\sin(x^2-4x+3)}$$ using only basic trigonometric identities from precalculus and the basic trigonometric limits $\lim\limits_{\theta\to0}\frac{\sin\theta}{\theta} = 1$ and $\lim\limits_{\theta\to0}\frac{1-\cos\theta}{\theta} = 0$, i.e. without using L'Hôpital's rule or Taylor expansions.

I've made some headway using the substitution $u = x-1$ after some general simplifying, but cannot get a clean result.

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We have by $y=x-1 \to 0$

$$\lim\limits_{x\to1}\frac{\tan(x^2-1)}{\sin(x^2-4x+3)}=\lim\limits_{y\to0}\frac{\tan(y(y+2))}{\sin(y(y-2))}$$

and

$$\frac{\tan(y(y+2))}{\sin(y(y-2))}=\frac{\tan(y(y+2))}{y(y+2)}\frac{y(y-2)}{\sin(y(y-2))}\frac{y(y+2)}{y(y-2)}$$

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Put $$a=x^2-1=(x-1)(x+1)$$ and $$b=x^2-4x+3=(x-1)(x-3).$$

observe that when $ x \to 1 $, $ a $ and $b $ go to zero. so

$$\lim_{x\to 1}\frac{\tan(b)}{b}\frac{a}{\sin(a)}\frac{b}{a}$$

$$=\lim_{x\to 1}\frac ba=\lim_{x\to 1}\frac{x-3}{x+1}=-1$$

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  • $\begingroup$ Since it is for precalculus, expliciting $a\to 0$ and $b\to 0$ would be great. $\endgroup$ – zwim Sep 28 '19 at 21:28
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    $\begingroup$ @zwim Thanks . I did what you told me to do. $\endgroup$ – hamam_Abdallah Sep 28 '19 at 22:50

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