Question

practice exercises
19–26. evaluating limits graphically sketch a graph of f and use it to
make a conjecture about the values of f(a), $\lim\limits_{x \to a^-} f(x)$, $\lim\limits_{x \to a^+} f(x)$, and
$\lim\limits_{x \to a} f(x)$ or state that they do not exist.

1. $f(x) = \

$$\begin{cases} x^2 + 1 & \\text{if } x \\leq -1 \\\\ 3 & \\text{if } x > -1 \\end{cases}$$

$; $a = -1$

Explanation:

Step1: Find \( f(-1) \)

For \( x = -1 \), we use the first part of the piece - wise function \( f(x)=x^{2}+1 \) (since \( - 1\leq - 1\)). Substitute \( x=-1 \) into \( x^{2}+1 \): \( (-1)^{2}+1=1 + 1=2 \). So \( f(-1) = 2 \).

Step2: Find \( \lim_{x

ightarrow - 1^{-}}f(x) \)
As \( x
ightarrow - 1^{-} \), \( x < - 1 \), so we use \( f(x)=x^{2}+1 \). We know that for the function \( y=x^{2}+1 \), as \( x
ightarrow - 1 \) from the left, we substitute \( x=-1 \) into \( x^{2}+1 \). \( \lim_{x
ightarrow - 1^{-}}(x^{2}+1)=(-1)^{2}+1 = 2 \).

Step3: Find \( \lim_{x

ightarrow - 1^{+}}f(x) \)
As \( x
ightarrow - 1^{+} \), \( x>-1 \), so we use \( f(x) = 3 \). The limit of a constant function \( y = 3 \) as \( x
ightarrow - 1^{+} \) is the constant itself. So \( \lim_{x
ightarrow - 1^{+}}f(x)=3 \).

Step4: Find \( \lim_{x

ightarrow - 1}f(x) \)
For the limit \( \lim_{x
ightarrow a}f(x) \) to exist, \( \lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x) \). Here, \( \lim_{x
ightarrow - 1^{-}}f(x) = 2 \) and \( \lim_{x
ightarrow - 1^{+}}f(x)=3 \). Since \( 2
eq3 \), \( \lim_{x
ightarrow - 1}f(x) \) does not exist.

Answer:

\( f(-1)=2 \), \( \lim_{x
ightarrow - 1^{-}}f(x)=2 \), \( \lim_{x
ightarrow - 1^{+}}f(x)=3 \), \( \lim_{x
ightarrow - 1}f(x) \) does not exist.