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.


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





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

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

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

  • $\begingroup$ Since it is for precalculus, expliciting $a\to 0$ and $b\to 0$ would be great. $\endgroup$ – zwim Sep 28 '19 at 21:28
  • 1
    $\begingroup$ @zwim Thanks . I did what you told me to do. $\endgroup$ – hamam_Abdallah Sep 28 '19 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.