Question

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Answer 1

$\displaystyle a)\,\, \lim\limits_{x\to 1}\frac{1}{(x-1)^2} = \frac{1}{(1-1)^2} = \frac{1}{(\pm 0)^2} =\frac{1}{+0} = +\infty$

$\displaystyle b)\,\, \lim\limits_{x\to -1}\frac{2}{(x+1)^2} = \frac{2}{(-1+1)^2} = \frac{2}{(\pm 0)^2} = \frac{2}{+0} = +\infty$

$\displaystyle c)\,\, \lim\limits_{x\to 2}\frac{3x-4}{x^2-4x+4} = \lim\limits_{x\to 2}\frac{3x-4}{(x-2)^2} = \frac{3\cdot 2-4}{(2-2)^2} = \frac{6-4}{(\pm 0)^2} = \frac{2}{+0} = +\infty$

$\displaystyle d)\,\, \lim\limits_{x\to 3}\frac{6x}{-x^2+6x-9} = \lim\limits_{x\to 2}\frac{6x}{-(x^2-6x+9)} = \lim\limits_{x\to 2}\frac{6x}{-(x-3)^2} =$

$\displaystyle = \frac{6\cdot 3 }{-(3-3)^2} = \frac{18}{-(\pm 0)^2} = \frac{18}{-(+0)} = \frac{18}{-0} = -\infty$

$\displaystyle e)\,\, \lim\limits_{x\to 0}\frac{3x+11}{x^2(x+1)} =\frac{3\cdot 0+11}{(\pm 0)^2(0+1)} = \frac{11}{(+0)\cdot 1} = \frac{11}{+0} = +\infty$

$\displaystyle f)\,\, \lim\limits_{x\to -1}\frac{4x+3}{1+2x+x^2} = \lim\limits_{x\to -1}\frac{4x+3}{(1+x)^2} = \frac{4\cdot (-1)+3}{(1-1)^2} = \frac{-4+3}{(\pm 0)^2} =$

$\displaystyle = \frac{-1}{+0} = -\infty$