Math Snap

STEP 1

Assumptions1. The function $f(x)$ is defined piecewise for $x \neq0$ as follows - $f(x) =4+x$ for $x<-$
- $f(x) = \frac{3}{}x$ for $0<x\leq$
- $f(x) =3+x-\frac{1}{}x^{}$ for $x>$
. We are asked to find all values of $a$ at which the limit of $f(x)$ as $x$ approaches $a$ exists.

STEP 2

For a limit to exist at a certain point $a$, the left-hand limit and the right-hand limit at that point must be equal. That is, $\lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x)$.

STEP 3

Let's first consider the points where the definition of the function changes, namely $x=-2$, $x=0$, and $x=2$.

STEP 4

For $x=-2$, we have to check if $\lim_{x \rightarrow -2^-} f(x) = \lim_{x \rightarrow -2^+} f(x)$.

STEP 5

Calculate the left-hand limit at $x=-2$ using the definition of $f(x)$ for $x<-2$.
$$\lim_{x \rightarrow -2^-} f(x) = \lim_{x \rightarrow -2^-} (4+x)$$

STEP 6

Calculate the right-hand limit at $x=-2$ using the definition of $f(x)$ for $0<x\leq2$.
$$\lim_{x \rightarrow -2^+} f(x) = \lim_{x \rightarrow -2^+} \frac{3}{2}x$$

STEP 7

Notice that the right-hand limit at $x=-2$ does not exist because the function is not defined for $x=-2$ in the second piece. Therefore, the limit at $x=-2$ does not exist.

STEP 8

Now, let's consider $x=0$. We have to check if $\lim_{x \rightarrow0^-} f(x) = \lim_{x \rightarrow0^+} f(x)$.

STEP 9

Calculate the left-hand limit at $x=$ using the definition of $f(x)$ for $x<-2$.
$$\lim_{x \rightarrow^-} f(x) = \lim_{x \rightarrow^-} (4+x)$$

STEP 10

Calculate the right-hand limit at $x=0$ using the definition of $f(x)$ for $0<x\leq2$.
$$\lim_{x \rightarrow0^+} f(x) = \lim_{x \rightarrow0^+} \frac{3}{2}x$$

STEP 11

Notice that the left-hand limit at $x=0$ does not exist because the function is not defined for $x=0$ in the first piece. Therefore, the limit at $x=0$ does not exist.

STEP 12

Now, let's consider $x=2$. We have to check if $\lim_{x \rightarrow2^-} f(x) = \lim_{x \rightarrow2^+} f(x)$.

STEP 13

Calculate the left-hand limit at $x=2$ using the definition of $f(x)$ for $0<x\leq2$.
$$\lim_{x \rightarrow2^-} f(x) = \lim_{x \rightarrow2^-} \frac{3}{2}x$$

STEP 14

Calculate the right-hand limit at $x=2$ using the definition of $f(x)$ for $x>2$.
$$\lim_{x \rightarrow2^+} f(x) = \lim_{x \rightarrow2^+} (3+x-\frac{}{2}x^{2})$$

STEP 15

Calculate the values of the left-hand and right-hand limits at $x=2$.
$$\lim_{x \rightarrow2^-} f(x) =3$$$$\lim_{x \rightarrow2^+} f(x) =3$$

STEP 16

Since the left-hand limit equals the right-hand limit at $x=2$, the limit at $x=2$ exists.

STEP 17

For all other values of $x$ not equal to $-2$, $0$, or $2$, the function is continuous and therefore the limit exists.

STEP 18

Therefore, the values of $a$ at which $\lim_{x \rightarrow a} f(x)$ exists are all $x$ such that $x \neq -2,0$, and the limit exists at $x=2$.

Expressing this in interval notation, the values of $a$ at which $\lim_{x \rightarrow a} f(x)$ exists are $(-\infty, -) \cup (-,) \cup (,] \cup (, +\infty)$.
The values of $a$ at which the limit of $f(x)$ as $x$ approaches $a$ exists are $(-\infty, -) \cup (-,) \cup (,] \cup (, +\infty)$.