a)

$\underset{x\to -1^{+}}{\lim }\frac{\sqrt{x+1}+1}{x^2+1},x\in \mathbb{R}$

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

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b) 

$\underset{x\to 4^{+}}{\lim }\frac{1}{\sqrt{x^2-16}+2}$  

założenia:

$\sqrt{x^2-16}+2\ne 0\wedge x^2-16\ge 0$

$\sqrt{x^2-16}\ne -2\wedge x^2\ge 16$

$x\in \mathbb{R}\wedge \left|x\right|\ge 4$    

$x\ge 4\vee x\le -4$  

$x\in (\infty ,-4⟩\cup ⟨4,\infty )$  

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$\underset{x\to 4^{+}}{\lim }\frac{1}{\sqrt{x^2-16}+2}=\frac{1}{\sqrt{4^2-16}+2}=\frac{1}{2}$  

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c)

$\underset{x\to 0^{+}}{\lim }\frac{x^2-x}{\sqrt{x}}$  

założenia:

$x>0$  

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$\underset{x\to 0^{+}}{\lim }\frac{x^2-x}{\sqrt{x}}\cdot \frac{\sqrt{x}}{\sqrt{x}}=$  

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

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

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

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d)

$\underset{x\to 2^{-}}{\lim }\frac{x-2}{\sqrt{4-x^2}}$  

założenia:

$4-x^2>0$

$x^2<4$

$\left|x\right|<2$

$x<2\wedge x>-2$

$x\in \left(-2,2\right)$      

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$\underset{x\to 2^{-}}{\lim }\frac{x-2}{\sqrt{4-x^2}}\cdot \frac{\sqrt{4-x^2}}{\sqrt{4-x^2}}=$

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

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

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

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

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