a)  $\underset{x\to -1^{+}}{\lim }\frac{\sqrt{x+1}+1}{x^2+1}=\frac{\sqrt{-1+1}+1}{{\left(-1\right)}^2+1}=\frac{1}{2}$  

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b)  $\underset{x\to 4^{-}}{\lim }\frac{4-x}{\sqrt{4-x}}=\underset{x\to 4^{-}}{\lim }\frac{4-x}{\sqrt{4-x}}\cdot \frac{\sqrt{4-x}}{\sqrt{4-x}}=$  

$=\underset{x\to 4^{-}}{\lim }\frac{\left(4-x\right)\cdot \sqrt{4-x}}{4-x}=\underset{x\to 4^{-}}{\lim }\sqrt{4-x}=4-4=0$  

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c)  $\underset{x\to 2^{+}}{\lim }\frac{x-2}{\sqrt{x^2-4}}=\underset{x\to 2^{+}}{\lim }\frac{x-2}{\sqrt{x^2-4}}\cdot \frac{\sqrt{x^2-4}}{\sqrt{x^2-4}}=\underset{x\to 2^{+}}{\lim }\frac{\left(x-2\right)\left(\sqrt{x^2-4}\right)}{x^2-4}=$  

$=\underset{x\to 2^{+}}{\lim }\frac{\left(x-2\right)\left(\sqrt{x^2-4}\right)}{\left(x-2\right)\left(x+2\right)}=\underset{x\to 2^{+}}{\lim }\frac{\sqrt{x^2-4}}{x+2}=\frac{\sqrt{2^2-4}}{2+2}=\frac{0}{4}=0$ $=\underset{x\to 2^{+}}{\lim }\frac{\left(x-2\right)\left(\sqrt{x^2-4}\right)}{\left(x-2\right)\left(x+2\right)}=\underset{x\to 2^{+}}{\lim }\frac{\sqrt{x^2-4}}{x+2}=\frac{\sqrt{2^2-4}}{2+2}=\frac{0}{4}=0$  

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d)  $\underset{x\to 0^{+}}{\lim }\frac{x-\sqrt{x}}{\sqrt{x}}=\underset{x\to 0^{+}}{\lim }\frac{x-\sqrt{x}}{\sqrt{x}}\cdot \frac{\sqrt{x}}{\sqrt{x}}=$  

$=\underset{x\to 0^{+}}{\lim }\frac{x\sqrt{x}-x}{x}=\underset{x\to 0^{+}}{\lim }\left(\sqrt{x}-1\right)=0-1=-1$  

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e)  $\underset{x\to -2^{-}}{\lim }\frac{x^2+5x+6}{2x^2-8}$  

 

$x^2+5x+6=0$  

$\Delta =5^2-4\cdot 1\cdot 6=25-24=1$  

$x_1=\frac{-5-1}{2}=\frac{-6}{2}=-3$  

$x_2=\frac{-5+1}{2}=\frac{-4}{2}=-2$  

$x^2+5x+6\iff \left(x+3\right)\left(x+2\right)$  

 

$2x^2-8=2\left(x^2-4\right)=2\left(x-2\right)\left(x+2\right)$  

 

$\underset{x\to -2^{-}}{\lim }\frac{x^2+5x+6}{2x^2-8}=\underset{x\to -2^{-}}{\lim }\frac{\left(x+3\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}=$  

$=\underset{x\to -2^{-}}{\lim }\frac{x+3}{2\left(x-2\right)}=\frac{-2+3}{2\left(-2-2\right)}=\frac{1}{2\cdot \left(-4\right)}=-\frac{1}{8}$  

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f)  $\underset{x\to -3^{+}}{\lim }\frac{5x+15}{x^2+x-6}$  

 

$x^2+x-6=0$  

$\Delta =1^2-4\cdot 1\cdot \left(-6\right)=1+24=25$  

$x_1=\frac{-1-5}{2}=\frac{-6}{2}=-3$  

$x_2=\frac{-1+5}{2}=\frac{4}{2}=2$  

$x^2+x-6\iff \left(x+3\right)\left(x-2\right)$  

 

$\underset{x\to -3^{+}}{\lim }\frac{5x+15}{x^2+x-6}=\underset{x\to -3^{+}}{\lim }\frac{5\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\underset{x\to -3^{+}}{\lim }\frac{5}{x-2}=\frac{5}{-3-2}=\frac{5}{-5}=-1$