Daje Naj Pilne Oblicz lim (x->2) (1/(4x-8)+1/(4-x^2). Powinno wyjść 1/16.

[latex]displaystyle lim_{x o2}left(dfrac{1}{4x-8}+dfrac{1}{4-x^2} ight)=\ lim_{x o2}left(dfrac{1}{4(x-2)}+dfrac{1}{(2-x)(2+x)} ight)=\ lim_{x o2}left(dfrac{1}{4(x-2)}-dfrac{1}{(x-2)(x+2)} ight)=\ lim_{x o2}left(dfrac{x+2}{4(x-2)(x+2)}-dfrac{4}{4(x-2)(x+2)} ight)=\ lim_{x o2}dfrac{x-2}{4(x-2)(x+2)}=\ lim_{x o2}dfrac{1}{4(x+2)}=\ dfrac{1}{4(2+2)}=\ dfrac{1}{4cdot4}=\dfrac{1}{16}[/latex]

[latex]\ lim_{x o 2} ( frac{1}{4(x-2)} -frac{1}{(x+2)(x-2)}) = lim_{x o 2} frac{x+2-4}{4(x-2)(x+2)} = \ \ lim_{x o 2} frac{1}{4(x+2)} = frac{1}{4*(2+2)} = frac{1}{16} [/latex]