$ \displaystyle\lim_{x \to 5^-} \left( \frac{1}{2} \right)^{\frac{1}{x-5}} = $
Cambio variabile. t = x - 5; inoltre se x → 5⁻ allora t → 0⁻
$ = \displaystyle\lim_{t \to 0^- } \left( \frac{1}{2} \right)^{\frac{1}{t}} = \left( \frac{1}{2} \right)^{-\infty} = +\infty $
nota: 1/2 < 1
LIM((1/2)^(1/(x - 5)))= ??
x-->5-
1/(x - 5) = t
LIM(1/(x - 5))=-∞
x-->5-
LIM((1/2)^t)=+∞
t---> -∞