Question

find the following limits if ( f(x) = \begin{cases} x^2 + 5 & x < 1 \\ \frac{1}{x - 1} & x > 1 end{cases} ) start by finishing the table. then, use the table to find the limits requested below. fill in the table with the exact decimal values of ( f(x) ).

( x )	0	0.5	0.9	0.99	1	1.0001	1.01	1.1	1.5

give the value of the following limits. give infinite limits where appropriate. type two lower case o’s, oo, for ( infty ). type a negative and two lower case o’s, -oo, for ( -infty ).

if the limit does not exist, type dne in the answer blank.

( lim_{x \to 1^-} f(x) = square )

( lim_{x \to 1^+} f(x) = square )

( lim_{x \to 1} f(x) = square )

question help:
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Explanation:

Step1: Calculate \( f(x) \) for \( x < 1 \)

For \( x = 0 \): \( f(0)=0^{2}+5 = 5 \)
For \( x = 0.5 \): \( f(0.5)=0.5^{2}+5 = 0.25 + 5 = 5.25 \)
For \( x = 0.9 \): \( f(0.9)=0.9^{2}+5 = 0.81 + 5 = 5.81 \)
For \( x = 0.99 \): \( f(0.99)=0.99^{2}+5 = 0.9801 + 5 = 5.9801 \)

Step2: Calculate \( f(x) \) for \( x > 1 \)

For \( x = 1.0001 \): \( f(1.0001)=\frac{1}{1.0001 - 1}=\frac{1}{0.0001}=10000 \)
For \( x = 1.01 \): \( f(1.01)=\frac{1}{1.01 - 1}=\frac{1}{0.01}=100 \)
For \( x = 1.1 \): \( f(1.1)=\frac{1}{1.1 - 1}=\frac{1}{0.1}=10 \)
For \( x = 1.5 \): \( f(1.5)=\frac{1}{1.5 - 1}=\frac{1}{0.5}=2 \)

Step3: Find \( \lim_{x

ightarrow1^{-}}f(x) \)
As \( x \) approaches \( 1 \) from the left (\( x < 1 \)), we use \( f(x)=x^{2}+5 \).
\( \lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}(x^{2}+5)=1^{2}+5 = 6 \)

Step4: Find \( \lim_{x

ightarrow1^{+}}f(x) \)
As \( x \) approaches \( 1 \) from the right (\( x > 1 \)), we use \( f(x)=\frac{1}{x - 1} \).
As \( x
ightarrow1^{+} \), \( x - 1
ightarrow0^{+} \), so \( \frac{1}{x - 1}
ightarrow+\infty \) (since the denominator approaches 0 from the positive side, the fraction goes to positive infinity).

Step5: Find \( \lim_{x

ightarrow1}f(x) \)
For the limit as \( x
ightarrow1 \) to exist, the left - hand limit and the right - hand limit must be equal. Since \( \lim_{x
ightarrow1^{-}}f(x) = 6 \) and \( \lim_{x
ightarrow1^{+}}f(x)=+\infty \), and \( 6
eq+\infty \), the limit \( \lim_{x
ightarrow1}f(x) \) does not exist (DNE).

Table Filling:

\( x \)	0	0.5	0.9	0.99	1	1.0001	1.01	1.1	1.5

Limits:

• \( \lim_{x

ightarrow1^{-}}f(x)=\boldsymbol{6} \)

• \( \lim_{x

ightarrow1^{+}}f(x)=\boldsymbol{\infty} \)

• \( \lim_{x

ightarrow1}f(x)=\boldsymbol{DNE} \)

Answer:

• Table values: \( f(0) = 5 \), \( f(0.5)=5.25 \), \( f(0.9)=5.81 \), \( f(0.99)=5.9801 \), \( f(1.0001)=10000 \), \( f(1.01)=100 \), \( f(1.1)=10 \), \( f(1.5)=2 \)
• \( \lim_{x

ightarrow1^{-}}f(x)=\boldsymbol{6} \)

• \( \lim_{x

ightarrow1^{+}}f(x)=\boldsymbol{\infty} \)

• \( \lim_{x

ightarrow1}f(x)=\boldsymbol{DNE} \)