Question

$\lim_{n \to 2} \frac{\sqrt{x^{2} +12} -4}{2-\sqrt{x^{3}-4 } }$

Answer 1

$\large\boxed{\begin{array}{l}\displaystyle\sf\lim_{x \to 2}\dfrac{(\sqrt{x^2+12}-4)}{(2-\sqrt{x^3-4})}\cdot\dfrac{(\sqrt{x^2+12}+4)}{(\sqrt{x^2+12}+4)}\cdot\dfrac{(2+\sqrt{x^3-4})}{(2+\sqrt{x^3-4})}\\\\\displaystyle\sf\lim_{x \to 2}\dfrac{[(\sqrt{x^2+12})^2-4^2]\cdot(2+\sqrt{x^3-4})}{[2^2-(\sqrt{x^3-4})^2]\cdot(\sqrt{x^2+12}+4)}\\\\\displaystyle\sf\lim_{x \to 2}\dfrac{[x^2+12-16]\cdot(2+\sqrt{x^3-4})}{[4-x^3+4]\cdot(\sqrt{x^2+12}+4)}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\lim_{x \to 2}\dfrac{(x^2-4)\cdot(2+\sqrt{x^3-4})}{(8-x^3)\cdot(\sqrt{x^2+12}+4)}\\\\\displaystyle\sf\lim_{x \to 2}\dfrac{(x-2)(x+2)\cdot(2+\sqrt{x^3-4})}{(2-x)\cdot(4+2x+x^2)\cdot(\sqrt{x^2+12}+4)}\\\\\displaystyle\sf-\lim_{x \to 2}\dfrac{\diagup\!\!\!\!\!\!(x-\diagup\!\!\!\!\!\!2)\cdot(x+2)\cdot(2+\sqrt{x^3-4})}{\diagup\!\!\!\!\!\!(x-\diagup\!\!\!\!\!\!2)\cdot(4+2x+x^2)\cdot(\sqrt{x^2+12}+4)}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf-\dfrac{(2+2)\cdot(2+\sqrt{2^3-4})}{(4+2\cdot2+2^2)\cdot(\sqrt{2^2+12}+4)}\\\\\sf -\dfrac{\diagdown\!\!\!\!\!\!4\cdot\diagdown\!\!\!\!\!\!4}{\diagdown\!\!\!\!\!\!12_3\cdot\diagdown\!\!\!\!\!8_2}=-\dfrac{1}{6}\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{x \to 2}\dfrac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}=-\dfrac{1}{6}}}}}\end{array}}$