[m]\lim \limits_{x \to 0-0} \frac{1}{2^{3/x} - 2} = \frac{1}{2^{-\infty} - 2} = \frac{1}{0 - 2} = -\frac{1}{2} = -0,5[/m]
[m]\lim \limits_{x \to 0+0} \frac{1}{2^{3/x} - 2} = \frac{1}{2^{+\infty} - 2} = \frac{1}{+\infty - 2} = \frac{1}{+\infty} = 0[/m]
При x = 3 разрыв 2 вида - уход в бесконечность.
[m]\lim \limits_{x \to 3-0} \frac{1}{2^{3/x} - 2} = \frac{1}{-(2^{1} - 2)} = \frac{1}{-(2 - 2)} = -\frac{1}{0} = -\infty[/m]
[m]\lim \limits_{x \to 3+0} \frac{1}{2^{3/x} - 2} = \frac{1}{2^{1} - 2} = \frac{1}{2- 2} = \frac{1}{0} = +\infty[/m]