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$(3 pts) For the function f(x)={[2x," if "x < 1],[-4x," if "x >= 1]:}, evaluate the left and right limits using the table shown below: x 2x x -4x 0.9 1.8 1.1 -4.4 0.99 1.98 1.01 -4.04 0.999 1.998 1.001 -4.004 0.9999 1.9998 1.0001 -4.0004 0.99999 1.99998 1.00001 -4.00004 a) lim_(x rarr1^(-))f(x) b) lim_(x rarr1^(+))f(x)$

$1$ . ( $3$ pts) For the function $f(x)=\left\{\begin{array}{cl}2 x & \text { if } x<1 \\ -4 x & \text { if } x \geq 1\end{array}\right.$ , evaluate the left and right limits using the table shown below: $\newline$ \begin{tabular}{cccl} $\newline$ $x$ & $2 x$ & $x$ & $-4 x$ \\ $\newline$ \hline $0$ . $9$ & $1$ . $8$ & $1$ . $1$ & $-4$ . $4$ \\ $\newline$ \hline $0$ . $99$ & $1$ . $98$ & $1$ . $01$ & $-4$ . $04$ \\ $\newline$ \hline $0$ . $999$ & $1$ . $998$ & $1$ . $001$ & $-4$ . $004$ \\ $\newline$ \hline $0$ . $9999$ & $1$ . $9998$ & $1$ . $0001$ & $-4$ . $0004$ \\ $\newline$ \hline $0$ . $99999$ & $1$ . $99998$ & $1$ . $00001$ & $-4$ . $00004$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ a) $\lim _{x \rightarrow 1^{-}} f(x)$ $\newline$ b) $\lim _{x \rightarrow 1^{+}} f(x)$

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Precalculus

Pascal's triangle and the Binomial Theorem

Full solution

1

. (

3

pts) For the function

f(x)=\left\{\begin{array}{cl}2 x & \text { if } x<1 \\ -4 x & \text { if } x \geq 1\end{array}\right.

, evaluate the left and right limits using the table shown below:

\newline

\begin{tabular}{cccl}

\newline

x

2 x

x

-4 x

\newline

\hline

0

9

1

8

1

1

-4

4

\newline

\hline

0

99

1

98

1

01

-4

04

\newline

\hline

0

999

1

998

1

001

-4

004

\newline

\hline

0

9999

1

9998

1

0001

-4

0004

\newline

\hline

0

99999

1

99998

1

00001

-4

00004

\newline

\hline

\newline

\end{tabular}

\newline

\lim _{x \rightarrow 1^{-}} f(x)

\newline

\lim _{x \rightarrow 1^{+}} f(x)

Find Left-Hand Limit: To find the left-hand limit as $x$ approaches $1$ , we look at the values of $f(x)$ when $x$ is just less than $1$ . According to the function definition, for $x < 1$ , $f(x) = 2x$ . We will use the values from the table to evaluate the limit.
Evaluate Left-Hand Limit: Looking at the table, as $x$ approaches $1$ from the left ( $x < 1$ ), the values of $f(x) = 2x$ are getting closer to $2$ . The sequence of $x$ values ( $0.9$ , $0.99$ , $0.999$ , $0.9999$ , $1$ $0$ ) is approaching $1$ , and the corresponding $1$ $2$ values ( $1$ $3$ , $1$ $4$ , $1$ $5$ , $1$ $6$ , $1$ $7$ ) are approaching $2$ .
Find Right-Hand Limit: Therefore, the left-hand limit of $f(x)$ as $x$ approaches $1$ is $2$ . This is because as $x$ gets closer to $1$ from the left, $2x$ gets closer to $2$ .
Evaluate Right-Hand Limit: To find the right-hand limit as $x$ approaches $1$ , we look at the values of $f(x)$ when $x$ is just greater than $1$ . According to the function definition, for $x \geq 1$ , $f(x) = -4x$ . We will use the values from the table to evaluate the limit.
Evaluate Right-Hand Limit: To find the right-hand limit as $x$ approaches $1$ , we look at the values of $f(x)$ when $x$ is just greater than $1$ . According to the function definition, for $x \geq 1$ , $f(x) = -4x$ . We will use the values from the table to evaluate the limit.Looking at the table, as $x$ approaches $1$ from the right ( $x > 1$ ), the values of $f(x) = -4x$ are getting closer to $1$ $1$ . The sequence of $x$ values ( $1$ $3$ , $1$ $4$ , $1$ $5$ , $1$ $6$ , $1$ $7$ ) is approaching $1$ , and the corresponding $f(x)$ values ( $f(x)$ $0$ , $f(x)$ $1$ , $f(x)$ $2$ , $f(x)$ $3$ , $f(x)$ $4$ ) are approaching $1$ $1$ .
Evaluate Right-Hand Limit: To find the right-hand limit as $x$ approaches $1$ , we look at the values of $f(x)$ when $x$ is just greater than $1$ . According to the function definition, for $x \geq 1$ , $f(x) = -4x$ . We will use the values from the table to evaluate the limit.Looking at the table, as $x$ approaches $1$ from the right ( $x > 1$ ), the values of $f(x) = -4x$ are getting closer to $1$ $1$ . The sequence of $x$ values ( $1$ $3$ , $1$ $4$ , $1$ $5$ , $1$ $6$ , $1$ $7$ ) is approaching $1$ , and the corresponding $f(x)$ values ( $f(x)$ $0$ , $f(x)$ $1$ , $f(x)$ $2$ , $f(x)$ $3$ , $f(x)$ $4$ ) are approaching $1$ $1$ .Therefore, the right-hand limit of $f(x)$ as $x$ approaches $1$ is $1$ $1$ . This is because as $x$ gets closer to $1$ from the right, $x$ $2$ gets closer to $1$ $1$ .