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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.

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5 Answers 5

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

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\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}

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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 $0$.

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1  
We have $\sin(x)=x+O(x^3)$ so you should use big $O$ instead of little $o$. –  Sami Ben Romdhane 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 $2$.

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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 $f'(1)=2$.

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