Question

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Answer 1

$\displaystyle a)\,\, \lim\limits_{x\to 1}\frac{4x-4}{9x^2-9} = \lim\limits_{x\to 1}\frac{4(x-1)}{9(x^2-1)} = \lim\limits_{x\to 1}\frac{4(x-1)}{9(x-1)(x+1)} =$

$\displaystyle = \lim\limits_{x\to 1}\frac{4}{9(x+1)} = \frac{4}{9 (1+1)} = \frac{4}{9\cdot 2} = \frac{2}{9}$

$\displaystyle b)\,\, \lim\limits_{x\to -1}\frac{x^2-1}{x^2+3x+2} = \lim\limits_{x\to -1}\frac{(x-1)(x+1)}{(x+1)(x+2)} = \lim\limits_{x\to -1}\frac{x-1}{x+2} =$

$\displaystyle = \frac{-1-1}{-1+2} = \frac{-2}{1} = -2$

$\displaystyle c)\,\, \lim\limits_{x\to 2}\frac{x^2-4}{x^2-3x+2} = \lim\limits_{x\to 2}\frac{(x-2)(x+2)}{(x-1)(x-2)} = \lim\limits_{x\to 2}\frac{x+2}{x-1} = \frac{2+2}{2-1} = 4$

$\displaystyle d)\,\, \lim\limits_{x\to 3}\frac{x^2-3x}{x^2-7x+12} = \lim\limits_{x\to 3}\frac{x(x-3)}{(x-3)(x-4)} = \lim\limits_{x\to 3}\frac{x}{x-4} = \frac{3}{3-4} = -3$

$\displaystyle e)\,\, \lim\limits_{x\to 2}\frac{(x-2)^2}{x^2-2x} = \lim\limits_{x\to 2}\frac{(x-2)^2}{x(x-2)} = \lim\limits_{x\to 2}\frac{x-2}{x} = \frac{2-2}{2} = 0$

$\displaystyle f)\,\, \lim\limits_{x\to -2}\frac{x^2+4x+4}{2x^2+4x} = \lim\limits_{x\to -2}\frac{(x+2)^2}{2x(x+2)} = \lim\limits_{x\to -2}\frac{x+2}{2x} = \frac{-2+2}{2\cdot (-2)} = 0$