Question

consider the following.
$f(x) = \

$$\begin{cases} e^x & \\text{if } x < 1 \\\\ x^3 & \\text{if } x \\geq 1 \\end{cases}$$

, \quad a = 1$
find the left-hand and right-hand limits at the given value of $a$.
$\lim\limits_{x \to 1^-} f(x) = \square$
$\lim\limits_{x \to 1^+} f(x) = 1$
explain why the function is discontinuous at the given number $a$.
since these limits are not equal , $\lim\limits_{x \to 1} f(x)$ does not exist and $f$ is therefore discontinuous at $1$.
sketch the graph of the function.

Explanation:

Step1: Find left - hand limit

For the left - hand limit as \(x
ightarrow1^{-}\), we use the part of the piece - wise function where \(x < 1\), which is \(f(x)=e^{x}\). The left - hand limit is \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}e^{x}\).
Since the exponential function \(y = e^{x}\) is continuous everywhere, we can directly substitute \(x = 1\) into \(e^{x}\). So \(\lim_{x
ightarrow1^{-}}e^{x}=e^{1}=e\).

Step2: Find right - hand limit (already partially correct)

For the right - hand limit as \(x
ightarrow1^{+}\), we use the part of the piece - wise function where \(x\geq1\), which is \(f(x)=x^{3}\). The right - hand limit is \(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}x^{3}\).
Since the power function \(y=x^{3}\) is continuous everywhere, we substitute \(x = 1\) into \(x^{3}\), so \(\lim_{x
ightarrow1^{+}}x^{3}=1^{3}=1\).

Answer:

\(\lim_{x
ightarrow1^{-}}f(x)=e\), \(\lim_{x
ightarrow1^{+}}f(x)=1\)