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Can any one find the following limits for me:

1- $\lim\limits_{x \to 0}\frac{x^2-1}{2x^2-x-1}$

2- $\lim\limits_{x \to 1} \frac{x^2-1}{2x^2-x-1}$

3- $\lim\limits_{x \to \infty} \frac{x^2-1}{2x^2-x-1}$

Thank you all.

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2  
Hi, normally you will get a more helpful answer if you show some effort yourself. Can you edit your thoughts into the question? –  Aryabhata Apr 12 '13 at 0:01

3 Answers 3

1) Just substitute.

2) Try to eliminate common factors between numerator and denominator.

3) Is the same as 2 (or I need new glasses).

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Hopefully the OP didn't mean the $\epsilon-\delta$ definition. –  Lord Soth Apr 12 '13 at 0:07
    
OP said if possible, and given the effort OP has shown, I would say it is impossible :-) –  Aryabhata Apr 12 '13 at 0:10
    
Hahaha, fair enough :) –  Lord Soth Apr 12 '13 at 0:10

Hint (huge):

$$\frac{x^2-1}{2x^2-x-1}=\frac{(x-1)(x+1)}{(x-1)(2x+1)}\ldots$$

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Using L'Hospital:

1- $$\lim\limits_{x \to 0}\frac{x^2-1}{2x^2-x-1}=\lim\frac{2x}{4x-1}=\frac{0}{-1}=0$$

2- $$\lim\limits_{x \to 1} \frac{x^2-1}{2x^2-x-1}=\lim\frac{2x}{4x-1}=\lim\frac{2}{3}=\frac{2}{3}$$

3- $$\lim\limits_{x \to \infty} \frac{x^2-1}{2x^2-x-1}=\lim\frac{2x}{4x-1}=\lim\frac{2}{4}=\frac{1}{2}.$$

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Thanks Iuli, but what about using analysis, I mean, maybe squees theorem with algebra of limits, or whatever? –  LoveMath Apr 12 '13 at 1:17

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