Question

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Answer 1

$\displaystyle a)\,\, \lim\limits_{x\to -\infty} \frac{2x+3}{-x+4} = \lim\limits_{x\to -\infty} \frac{x\left(2+\frac{3}{x}\right)}{x\left(-1+\frac{4}{x}\right)} = \lim\limits_{x\to -\infty} \frac{2+\frac{3}{x}}{1-\frac{4}{x}} = \frac{2+\frac{3}{\infty}}{1-\frac{4}{\infty}} =$

$\displaystyle = \frac{2+0}{1-0} = 2$

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

$\displaystyle = \frac{\frac{4}{\infty^2}-1}{2+\frac{1}{\infty}+\frac{1}{\infty^2}} = \frac{0-1}{2+0+0} = -\frac{1}{2}$

$\displaystyle c)\,\, \lim\limits_{x\to -\infty} \frac{-2x+11}{6x-11} = \lim\limits_{x\to -\infty} \frac{x\left(-2+\frac{11}{x}\right)}{x\left(6-\frac{11}{x}\right)} = \lim\limits_{x\to -\infty} \frac{-2+\frac{11}{x}}{6-\frac{11}{x}} =$

$\displaystyle = \frac{-2+\frac{11}{-\infty}}{6-\frac{11}{-\infty}} = \frac{-2+0}{6-0} = -\frac{2}{6} = -\frac{1}{3}$

$\displaystyle d)\,\, \lim\limits_{x\to -\infty} \frac{2-x^2}{3x^2+4x+11} = \lim\limits_{x\to -\infty} \frac{x^2\left(\frac{2}{x^2}-1\right)}{x^2\left(3+\frac{4}{x}+\frac{11}{x^2}\right)} = \lim\limits_{x\to -\infty} \frac{\frac{2}{x^2}-1}{3+\frac{4}{x}+\frac{11}{x^2}} =$

$\displaystyle = \frac{\frac{2}{(-\infty)^2}-1}{3+\frac{4}{-\infty}+\frac{11}{(-\infty)^2}} = \frac{0-1}{3+0+0} = -\frac{1}{3}$

$\displaystyle e)\,\, \lim\limits_{x\to +\infty} \frac{3x^2+6x+3}{2x+1} = \lim\limits_{x\to +\infty} \frac{x^2\left(3+\frac{6}{x}+\frac{3}{x^2}\right)}{x\left(2+\frac{1}{x}\right)} = \lim\limits_{x\to +\infty} \frac{x(3+\frac{6}{x}+\frac{3}{x^2})}{2+\frac{1}{x}} =$

$\displaystyle = \frac{\infty(3+\frac{6}{\infty}+\frac{3}{\infty^2})}{2+\frac{1}{\infty}} = \frac{\infty(3+0+0)}{2+0} = \frac{\infty\cdot 3}{2} = \infty$

$\displaystyle f)\,\, \lim\limits_{x\to -\infty} \frac{6x^2-3x+11}{2x+6} = \lim\limits_{x\to +\infty} \frac{x^2\left(6-\frac{3}{x}+\frac{11}{x^2}\right)}{x\left(2+\frac{6}{x}\right)} = \lim\limits_{x\to +\infty} \frac{x(6-\frac{3}{x}+\frac{11}{x^2})}{2+\frac{6}{x}} =$

$\displaystyle = \frac{-\infty(6-\frac{3}{-\infty}+\frac{11}{(-\infty)^2})}{2+\frac{6}{-\infty}} = \frac{-\infty(3+0+0)}{2+0} = \frac{-\infty\cdot 3}{2} = -\infty$

$\displaystyle g)\,\, \lim\limits_{x\to +\infty} \frac{3x-2}{4x^2+6x+1} = \lim\limits_{x\to +\infty} \frac{x\left(3-\frac{2}{x}\right)}{x^2\left(4+\frac{6}{x}+\frac{1}{x^2}\right)} = \lim\limits_{x\to +\infty} \frac{3-\frac{2}{x}}{x\left(4+\frac{6}{x}+\frac{1}{x^2}\right)}=$

$\displaystyle = \frac{3-\frac{2}{\infty}}{\infty\left(4+\frac{6}{\infty}+\frac{1}{\infty^2}\right)} = \frac{3-0}{\infty(4+0+0)} = \frac{3}{\infty\cdot 4} = 0$

$\displaystyle h)\,\, \lim\limits_{x\to \infty} \frac{2x}{(x^2+1)(x-1)^2} = \lim\limits_{x\to \infty} \frac{2x}{x^2\left(1+\frac{1}{x^2}\right)\left[x\left(1-\frac{1}{x}\right)\right]^2} =$

$\displaystyle = \lim\limits_{x\to \infty} \frac{2x}{x^2\left(1+\frac{1}{x^2}\right)x^2\left(1-\frac{1}{x}\right)^2} = \lim\limits_{x\to \infty} \frac{2x}{x^4\left(1+\frac{1}{x^2}\right)\left(1-\frac{1}{x}\right)^2} =$

$\displaystyle = \lim\limits_{x\to \infty} \frac{2}{x^3\left(1+\frac{1}{x^2}\right)\left(1-\frac{1}{x}\right)^2} = \frac{2}{\infty^3\left(1+\frac{1}{\infty^2}\right)\left(1-\frac{1}{\infty}\right)^2} = \frac{2}{\infty^3(1+0)(1-0)^2} =$

$\displaystyle = \frac{2}{\infty^3\cdot 1\cdot 1^2} = \frac{2}{\infty} = 0$