Let f be defined by f(x)= 4−x2 & if x 1 2+x2 & if x > 1. Find each of the following limits if they exist: (i) x 1^- f(x) (ii) x 1^+ f(x) (iii) _x 1 f(x)

The function is defined at $x=-3$.

For $x \neq -3$, we can simplify the expression:

Now, find the limit:

The limit exists.

Since $\lim_{x \to -3} f(x) = -5$ and $f(-3) = -5$, we have $\lim_{x \to -3} f(x) = f(-3)$.
For all other values of $x$, the function is a polynomial (after simplification), which is continuous.
Therefore, the function $f(x)$ is continuous everywhere.
There are no values of $x$ for which $f(x)$ is not continuous.

The function is defined at $x=-3$.

As shown in part (a), for $x \neq -3$, $\frac{x^2+x-6}{x+3} = x-2$.

The limit exists.

Since $\lim_{x \to -3} h(x) = -5$ and $h(-3) = 1$, we have $\lim_{x \to -3} h(x) \neq h(-3)$.
Therefore, the function $h(x)$ is not continuous at $\boxed*{x=-3}*$ because the limit of the function as $x$ approaches $-3$ does not equal the function's value at $x=-3$.

The function is defined at $x=-3$.

Consider the one-sided limits:

Since the left-hand limit and the right-hand limit are not equal (and both are infinite), the limit $\lim_{x \to -3} g(x)$ does not exist.
Therefore, the function $g(x)$ is not continuous at $\boxed*{x=-3}*$ because the limit of the function as $x$ approaches $-3$ does not exist.

The expression $\frac{|x|}{x}$ involves division by $x$. If $x=0$, the denominator is zero.

The function $f(x)$ is undefined at $x=0$.
Therefore, the function $f(x)$ is not continuous at $\boxed*{x=0}*$ because the function is undefined at this point.