Let's solve the given problems by evaluating the limits as specified.

17. $\lim_{{x \to 2}} \frac{1}{{x^2 - 4}}$

The function is $\frac{1}{{x^2 - 4}}$, which can be rewritten as $\frac{1}{{(x-2)(x+2)}}$.

As $x \to 2^+$:

• As $x$ approaches 2 from the right, $(x-2)$ is positive but close to zero, and $(x+2)$ is positive. Therefore, $\frac{1}{(x-2)(x+2)}$ approaches $+\infty$.

As $x \to 2^-$:

• As $x$ approaches 2 from the left, $(x-2)$ is negative but close to zero, and $(x+2)$ is positive. Therefore, $\frac{1}{(x-2)(x+2)}$ approaches $-\infty$.

As $x \to -2^+$:

• As $x$ approaches $-2$ from the right, $(x+2)$ is positive but close to zero, and $(x-2)$ is negative. Therefore, $\frac{1}{(x-2)(x+2)}$ approaches $-\infty$.

As $x \to -2^-$:

• As $x$ approaches $-2$ from the left, $(x+2)$ is negative and close to zero, and $(x-2)$ is negative. Therefore, (\