a)

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

założenia:

${\left(x-2\right)}^2\ne 0\wedge x-4\ne 0$

$x\ne 2\wedge x\ne 4$    

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Zauważamy, że 

${\left(x-2\right)}^2>0\text{dla}x\in \left(-\infty ,2\right)\cup \left(2,\infty \right)$  

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

zatem:

$\underset{x\to 2^{+}}{\lim }\left(\frac{1}{{\left(x-2\right)}^2}-\frac{1}{x-4}\right)=\frac{1}{0^{+}}-\frac{1}{-2}=$  

$=\infty +\frac{1}{2}=\infty$  

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b)

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

założenia:

${\left(x-2\right)}^2\ne 0\wedge x^2-4\ne 0$

$x-2\ne 0\wedge x^2\ne 4$

$x\ne 2\wedge x\ne -2$    

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Zauważamy, że

${\left(x-2\right)}^2>0\text{dla}x\in \left(-\infty ,2\right)\cup \left(2,\infty \right)$  

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

$x^2-4>0\text{dla}x\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$  

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

zatem:

$\underset{x\to 2^{+}}{\lim }\left(\frac{1}{{\left(x-2\right)}^2}-\frac{1}{x^2-4}\right)\stackrel{\left[\infty -\infty \right]}{=}$  

Doprowadzamy ułamki do wspólnego mianownika $\left(x-2\right)\left(x-2\right)\left(x+2\right)={\left(x-2\right)}^2\left(x+2\right)$

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

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

$=\underset{x\to 2^{+}}{\lim }\frac{4}{{\left(x-2\right)}^2\left(x+2\right)}=\frac{4}{0^{+}\cdot 4}=\frac{1}{0^{+}}=\infty$