Evaluate the following limits:

$\displaystyle \text{Question 1: } \lim \limits_{x \to 1} \frac{x^2+1}{x+1}$

$\displaystyle \lim \limits_{x \to 1} \frac{x^2+1}{x+1} = \frac{1^2+1}{1+1} = 1$

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$\displaystyle \text{Question 2: } \lim \limits_{x \to 0} \frac{2x^2+3x+4}{x^2 + 3x + 2}$

$\displaystyle \lim \limits_{x \to 0} \frac{2x^2+3x+4}{x^2 + 3x + 2} = \frac{2(0)^2+3(0)+4}{(0)^2 + 3(0) + 2} = \frac{4}{2} = 2$

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$\displaystyle \text{Question 3: } \lim \limits_{x \to 3} \frac{\sqrt{2x+3}}{x+3}$

$\displaystyle \lim \limits_{x \to 3} \frac{\sqrt{2x+3}}{x+3} = \frac{\sqrt{2(3)+3}}{3+3} = \frac{3}{6} = \frac{1}{2}$

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$\displaystyle \text{Question 4: } \lim \limits_{x \to 1} \frac{\sqrt{x+8}}{\sqrt{x}}$

$\displaystyle \lim \limits_{x \to 1} \frac{\sqrt{x+8}}{\sqrt{x}} = \frac{\sqrt{1+8}}{\sqrt{1}} = 3$

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$\displaystyle \text{Question 5: } \lim \limits_{x \to a} \frac{\sqrt{x}+\sqrt{a}}{x+a}$

$\displaystyle \lim \limits_{x \to a} \frac{\sqrt{x}+\sqrt{a}}{x+a} = \frac{\sqrt{a}+\sqrt{a}}{a+a} = \frac{2\sqrt{a}}{2a} = \frac{1}{\sqrt{a}}$

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$\displaystyle \text{Question 6: } \lim \limits_{x \to 1} \frac{1 + ( x-1)^2}{1+x^2}$

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

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$\displaystyle \text{Question 7: } \lim \limits_{x \to 0} \frac{x^{2/3}-9}{x-27}$

$\displaystyle \lim \limits_{x \to 0} \frac{x^{2/3}-9}{x-27} = \frac{0^{2/3}-9}{0-27} = \frac{1}{3}$

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$\displaystyle \text{Question 8: } \lim \limits_{x \to 0} 9$

$\displaystyle \lim \limits_{x \to 0} 9 = 9$

$\displaystyle f(x) = 9$ is a constant function. Its value does not depend on $\displaystyle x.$

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$\displaystyle \text{Question 9: } \lim \limits_{x \to 2} (3-x)$

$\displaystyle \lim \limits_{x \to 2} (3-x) = (3-2) = 1$

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$\displaystyle \text{Question 10: } \lim \limits_{x \to -1} (4x^2+2)$

$\displaystyle \lim \limits_{x \to -1} (4x^2+2) = (4(-1)^2+2) = 4+2 = 6$

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$\displaystyle \text{Question 11: } \lim \limits_{x \to -1} \frac{x^3-3x+1}{x-1}$

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

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$\displaystyle \text{Question 12: } \lim \limits_{x \to 0} \frac{3x+1}{x+3}$

$\displaystyle \lim \limits_{x \to 3} \frac{3x+1}{x+3} = \frac{3(0)+1}{0+3} = \frac{1}{3}$

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$\displaystyle \text{Question 13: } \lim \limits_{x \to 3} \frac{x^2-9}{x+2}$

$\displaystyle\lim \limits_{x \to 0} \frac{x^2-9}{x+2} = \frac{(3)^2-9}{3+2} = \frac{9-9}{5} = 0$
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$\displaystyle \text{Question 14: } \lim \limits_{x \to 0} \frac{ax+b}{cx+d} , \ \ d \neq 0$
$\displaystyle \lim \limits_{x \to 0} \frac{ax+b}{cx+d} = \frac{a(0)+b}{c(0)+d} = \frac{b}{d}$