Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to simplify this limit? 1 hour already and I haven't figured out a way yet. $$\lim_{x\to 1} \frac{\sin(x^2-1)}{x-1}$$

Withou using l'hôpital's rule, please.

share|cite|improve this question

Hint: $$\frac{\sin(x^2-1)}{x-1} = (x+1)\frac{\sin(x^2-1)}{x^2-1}$$

share|cite|improve this answer

\begin{align} \lim_{x\to 1} \dfrac{\sin(x^2-1)}{x-1}=&\lim_{x\to 1} \dfrac{\sin(x^2-1)}{x-1}\times\dfrac{x+1}{x+1}\\=&\lim_{x^2-1\to 0} \dfrac{\sin(x^2-1)}{x^2-1}\times\lim_{x\to1}(x+1)\\=&1\times(1+1)\\=&2 \end{align}

share|cite|improve this answer

As $x$ approaches $1$, we have by Taylor's theorem that $\sin(x^2 - 1) = x^2 - 1 + O(x^2 -1)^3$, so we get that

$$\lim_{x\to 1} \frac{\sin(x^2-1)}{x-1} = \lim_{x\to 1} \frac{x^2-1+ O(x^2 -1)^3}{x-1}$$ $$= \lim_{x\to 1} x+1 +O(x^2 -1)^2(x+1) = 2$$

as the big-oh term goes to $0$.

share|cite|improve this answer
We have $\sin(x)=x+O(x^3)$ so you should use big $O$ instead of little $o$. – user63181 Jun 9 '13 at 21:31
@SamiBenRomdhane Indeed; thanks for pointing this out! – Andrew D Jun 9 '13 at 21:32

It was just stop to think for a while that the answer came to me, dividing and multiplying the nominator by $x^2-1$ will give me $1$ from the fundamental trig limit times $x^2-1$. So it will be $\frac{x^2-1}{x-1}$, factoring the nominator and dividing it by $x-1$ will have just $x+1$, which is equal to $2$.

share|cite|improve this answer
Provided you can assume the result that $\lim_{x\to 0} \frac{\sin x}{x} = 1$, that's fine. – Andrew D Jun 9 '13 at 21:12
Oh, yes, I can. My teacher just doesn't allow the use of l'Hôpital's rule, at least not till we study derivatives. Anyway, she never taught us anything about Taylor's theorem. – Luan Cristian Thums Jun 9 '13 at 21:16
That makes sense if you haven't studied derivatives yet, although how you can prove the fundamental trig limit without some use of calculus is beyond me (although this is a minor point). – Andrew D Jun 9 '13 at 21:21

$$ \lim_{x\to 1}\frac{\sin(x^2-1)}{x-1}=\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=f'(1) $$ where $f(x)=\sin(x^2-1)$. The derivative is found using the chain rule: $$ f'(x)=\cos(x^2-1)\cdot 2x $$ and hence $f'(1)=2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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