Question

in the following problems you will find the limit of a function numerically (i.e. from a table). find the value of the limit from the functional values in the table. give infinite limits where appropriate. type two lower case o’s, oo, for ∞. type a negative and two lower case o’s, -oo, for −∞. if the limit does not exist, type dne in the answer blank. a) table with ( x ) values: 2.5, 2.9, 2.99, 2.999, 3, 3.001, 3.01, 3.1, 3.5 and function ( \frac{1}{x-3} ) with values: -2, -10, -100, -1000, und, 1000, 100, 10, 2 then three limit expressions: ( limlimits_{x \to 3^-} \frac{1}{x-3} = ), ( limlimits_{x \to 3^+} \frac{1}{x-3} = ), ( limlimits_{x \to 3} \frac{1}{x-3} = ). b) table with ( x ) values: -5, -4.5, -4.1, -4.01, -4, -3.99, -3.9, -3.5, -3 and function ( \frac{1}{(x+4)^2} ) with values: 1, 4, 100, 10000, und, 10000, 100, 4, 1 then three limit expressions: ( limlimits_{x \to -4^-} \frac{1}{(x+4)^2} = ), ( limlimits_{x \to -4^+} \frac{1}{(x+4)^2} = ), ( limlimits_{x \to -4} \frac{1}{(x+4)^2} = ).

Explanation:

Part A

1. Left - hand limit as \(x

ightarrow3^{-}\)

Step1: Analyze the table for \(x

ightarrow3^{-}\)
As \(x\) approaches \(3\) from the left (values like \(2.9, 2.99, 2.999\)), the values of \(\frac{1}{x - 3}\) are \(- 10,-100,-1000\) which are getting more and more negative (approaching \(-\infty\)).
So, \(\lim_{x
ightarrow3^{-}}\frac{1}{x - 3}=-\infty\)

2. Right - hand limit as \(x

ightarrow3^{+}\)

Step1: Analyze the table for \(x

ightarrow3^{+}\)
As \(x\) approaches \(3\) from the right (values like \(3.001, 3.01, 3.1\)), the values of \(\frac{1}{x - 3}\) are \(1000,100,10\) which are getting more and more positive (approaching \(+\infty\)).
So, \(\lim_{x
ightarrow3^{+}}\frac{1}{x - 3}=\infty\)

3. Overall limit as \(x

ightarrow3\)

Step1: Compare left and right limits

Since \(\lim_{x
ightarrow3^{-}}\frac{1}{x - 3}=-\infty\) and \(\lim_{x
ightarrow3^{+}}\frac{1}{x - 3}=\infty\), and the left - hand limit and right - hand limit are not equal, the limit \(\lim_{x
ightarrow3}\frac{1}{x - 3}\) does not exist (DNE).

Part B

1. Left - hand limit as \(x

ightarrow - 4^{-}\)

Step1: Analyze the table for \(x

ightarrow - 4^{-}\)
As \(x\) approaches \(-4\) from the left (values like \(-4.1,-4.01\)), the values of \(\frac{1}{(x + 4)^{2}}\) are \(100,10000\) which are getting larger and larger (approaching \(+\infty\)).
So, \(\lim_{x
ightarrow - 4^{-}}\frac{1}{(x + 4)^{2}}=\infty\)

2. Right - hand limit as \(x

ightarrow - 4^{+}\)

Step1: Analyze the table for \(x

ightarrow - 4^{+}\)
As \(x\) approaches \(-4\) from the right (values like \(-3.99,-3.9\)), the values of \(\frac{1}{(x + 4)^{2}}\) are \(10000,100\) which are getting larger and larger (approaching \(+\infty\)).
So, \(\lim_{x
ightarrow - 4^{+}}\frac{1}{(x + 4)^{2}}=\infty\)

3. Overall limit as \(x

ightarrow - 4\)

Step1: Compare left and right limits

Since \(\lim_{x
ightarrow - 4^{-}}\frac{1}{(x + 4)^{2}}=\infty\) and \(\lim_{x
ightarrow - 4^{+}}\frac{1}{(x + 4)^{2}}=\infty\), the limit \(\lim_{x
ightarrow - 4}\frac{1}{(x + 4)^{2}}=\infty\)

Final Answers

Part A

\(\lim_{x
ightarrow3^{-}}\frac{1}{x - 3}=\boldsymbol{-\infty}\)

\(\lim_{x
ightarrow3^{+}}\frac{1}{x - 3}=\boldsymbol{\infty}\)

\(\lim_{x
ightarrow3}\frac{1}{x - 3}=\boldsymbol{DNE}\)

Part B

\(\lim_{x
ightarrow - 4^{-}}\frac{1}{(x + 4)^{2}}=\boldsymbol{\infty}\)

\(\lim_{x
ightarrow - 4^{+}}\frac{1}{(x + 4)^{2}}=\boldsymbol{\infty}\)

\(\lim_{x
ightarrow - 4}\frac{1}{(x + 4)^{2}}=\boldsymbol{\infty}\)

Answer:

Part A

1. Left - hand limit as \(x

ightarrow3^{-}\)

Step1: Analyze the table for \(x

ightarrow3^{-}\)
As \(x\) approaches \(3\) from the left (values like \(2.9, 2.99, 2.999\)), the values of \(\frac{1}{x - 3}\) are \(- 10,-100,-1000\) which are getting more and more negative (approaching \(-\infty\)).
So, \(\lim_{x
ightarrow3^{-}}\frac{1}{x - 3}=-\infty\)

2. Right - hand limit as \(x

ightarrow3^{+}\)

Step1: Analyze the table for \(x

ightarrow3^{+}\)
As \(x\) approaches \(3\) from the right (values like \(3.001, 3.01, 3.1\)), the values of \(\frac{1}{x - 3}\) are \(1000,100,10\) which are getting more and more positive (approaching \(+\infty\)).
So, \(\lim_{x
ightarrow3^{+}}\frac{1}{x - 3}=\infty\)

3. Overall limit as \(x

ightarrow3\)

Step1: Compare left and right limits

Since \(\lim_{x
ightarrow3^{-}}\frac{1}{x - 3}=-\infty\) and \(\lim_{x
ightarrow3^{+}}\frac{1}{x - 3}=\infty\), and the left - hand limit and right - hand limit are not equal, the limit \(\lim_{x
ightarrow3}\frac{1}{x - 3}\) does not exist (DNE).

Part B

1. Left - hand limit as \(x

ightarrow - 4^{-}\)

Step1: Analyze the table for \(x

ightarrow - 4^{-}\)
As \(x\) approaches \(-4\) from the left (values like \(-4.1,-4.01\)), the values of \(\frac{1}{(x + 4)^{2}}\) are \(100,10000\) which are getting larger and larger (approaching \(+\infty\)).
So, \(\lim_{x
ightarrow - 4^{-}}\frac{1}{(x + 4)^{2}}=\infty\)

2. Right - hand limit as \(x

ightarrow - 4^{+}\)

Step1: Analyze the table for \(x

ightarrow - 4^{+}\)
As \(x\) approaches \(-4\) from the right (values like \(-3.99,-3.9\)), the values of \(\frac{1}{(x + 4)^{2}}\) are \(10000,100\) which are getting larger and larger (approaching \(+\infty\)).
So, \(\lim_{x
ightarrow - 4^{+}}\frac{1}{(x + 4)^{2}}=\infty\)

3. Overall limit as \(x

ightarrow - 4\)

Step1: Compare left and right limits

Since \(\lim_{x
ightarrow - 4^{-}}\frac{1}{(x + 4)^{2}}=\infty\) and \(\lim_{x
ightarrow - 4^{+}}\frac{1}{(x + 4)^{2}}=\infty\), the limit \(\lim_{x
ightarrow - 4}\frac{1}{(x + 4)^{2}}=\infty\)

Final Answers

Part A

\(\lim_{x
ightarrow3^{-}}\frac{1}{x - 3}=\boldsymbol{-\infty}\)

\(\lim_{x
ightarrow3^{+}}\frac{1}{x - 3}=\boldsymbol{\infty}\)

\(\lim_{x
ightarrow3}\frac{1}{x - 3}=\boldsymbol{DNE}\)

Part B

\(\lim_{x
ightarrow - 4^{-}}\frac{1}{(x + 4)^{2}}=\boldsymbol{\infty}\)

\(\lim_{x
ightarrow - 4^{+}}\frac{1}{(x + 4)^{2}}=\boldsymbol{\infty}\)

\(\lim_{x
ightarrow - 4}\frac{1}{(x + 4)^{2}}=\boldsymbol{\infty}\)