Indeterminate form 1 raised to infinity

Let's suppose that $lim_{x \to + \infty} f (x) = 1$ and $lim_{x \to + \infty} g (x) = \pm \infty$ , then we have that $lim_{x \to + \infty} f (x)^{g (x)} = 1^{\pm \infty}$ and we have again an indeterminate form.

The basic problem of this indeterminate form is to know from where $f (x)$ tends to one (right or left) and what function reaches its limit more rapidly.

To solve this limit we will use the following two formulae, depending on which it is more convenient. These convert the indeterminate form to one that we can solve.

The two formulae are the following: If $lim_{x \to + \infty} f (x) = 1$ and $lim_{x \to + \infty} g (x) = \pm \infty$ then,

$lim_{x \to + \infty} f (x)^{g (x)} = e^{(lim_{x \to + \infty} (f (x) - 1) \cdot g (X))}$
$lim_{x \to + \infty} f (x)^{g (x)} = e^{(lim_{x \to + \infty} g (x) \cdot \ln f (x))}$

Let's see some examples:

Example

1) $lim_{x \to + \infty} (\frac{1}{1 + x^{2}})^{2} x = e^{lim_{x \to + \infty} (\frac{1}{1 + x^{2}} - 1) \cdot 2 x} =$

$= e^{lim_{x \to + \infty} (\frac{1}{1 + x^{2}} - \frac{1 + x^{2}}{1 + x^{2}}) \cdot 2 x} = e^{lim_{x \to + \infty} \frac{- x^{2}}{1 + x^{2}} \cdot 2 x} =$

$= e^{lim_{x \to + \infty} \frac{- 2 x^{3}}{1 + x^{2}}} = e^{- \infty} = 0$

2) $lim_{x \to + \infty} (\frac{1}{1 + x^{2}})^{2} x = e^{lim_{x \to + \infty} 2 x \cdot (\frac{1}{1 + x^{2}})} =$

$= e^{lim_{x \to + \infty} - 2 \cdot \ln (1 + x^{2})} = e^{- \infty} = 0$

3) $lim_{x \to + \infty} (1 - \frac{2}{x^{2}})^{x^{2}} = e^{lim_{x \to + \infty} (1 - \frac{2}{x^{2}} - 1) \cdot x^{2}} =$

$= e^{lim_{x \to + \infty} (- \frac{2 x^{2}}{x^{2}})} = e^{- 2} = \frac{1}{e^{2}}$

4) $lim_{x \to + \infty} (1 + \frac{1}{2^{x}})^{x} = e^{lim_{x \to + \infty} (1 + \frac{1}{2^{x}} - 1) \cdot x} = e^{lim_{x \to + \infty} \frac{x}{2^{x}}} = e^{0} = 1$