Como realizar los siguientes limites de forma indeterminada.

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Hola Elkin!

3.- Se hace aplicando la igualdad ciclotómica:

$$\begin{align}&a^3-b^3=(a-b)(a^2+ab+b^2)\\&\\&a= \sqrt[3] x \Rightarrow a^3=x\\&b=1 \Rightarrow b^3=1\\&\\&a-b=\frac{a^3-b^3}{a^2+ab+b^2}\\&\\&\lim_{x \to 1} \frac { \sqrt [3] x -1}{x-1}= \frac 0 0= \lim_{x \to 1} \frac {x-1}{( \sqrt [3] {x^2} + \sqrt[3] x +1)(x-1)}=\\&\\& \lim_{x \to 1} \frac {1}{( \sqrt [3] {x^2} + \sqrt[3] x +1)}= \frac {1}{1+1+1}= \frac 1 3\\&\\&b)\\&\lim _{n \to 0} \frac  { \sqrt{5+n} -\sqrt {5}} { \sqrt {2n}}= \frac 0 0=\\&\\&\lim _{n \to 0} \frac  { (\sqrt{5+n} -\sqrt {5})(\sqrt{5+n} +\sqrt {5})} { \sqrt {2n} (\sqrt{5+n} +\sqrt {5})}=\\&\\&(A-B)(A+B)=A^2-B^2\\&\\&\lim _{n \to 0} \frac  { 5+n-5} { \sqrt {2n} (\sqrt{5+n} +\sqrt {5})}=\\&\\&\lim _{n \to 0} \frac  {n} { \sqrt {2n} (\sqrt{5+n} +\sqrt {5})}= \frac 0 0=\\&\\& \frac 1 {\sqrt 2}  \lim _{n \to 0} \frac  {n} { \sqrt {n} (\sqrt{5+n} +\sqrt {5})}=\\&\\& \frac 1 {\sqrt 2}  \lim _{n \to 0} \frac  {\sqrt n} { (\sqrt{5+n} +\sqrt {5})}= \frac 1 {\sqrt 2} · \frac 0 {2 \sqrt 5}=0\\&\\&\\&5)\\&\lim_{x \to 2} \frac {(x-2)^2}{x^2-4}= \frac 0 0= \lim_{x \to 2} \frac {(x-2)^2}{(x+2)(x-2)}=\\&\\&\lim_{x \to 2} \frac {(x-2)}{(x+2)}= \frac 0 4=0\end{align}$$

Saludos

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