Question

We wish to show that $\lim_{x \to 2} 4x + 5 = 13$ If we give an $\epsilon$, find the value of $\delta$ to use. Let $\epsilon = .04$ to find $\delta$ First set $| (4x + 5) - 13 | < .04$ Based on the absolute value, this means _ $< 4x - 8 <$ _ and _ $< 4x <$ _ and _ $< x <$ _ and This means we set $\delta = $ (round to four decimal places) So for any $x$ such that $|x - 2| <$ _ , we know that $| (4x + 5) - 13 | <$ _ Please answer all parts of the question.

          We wish to show that $\lim_{x \to 2} 4x + 5 = 13$

If we give an $\epsilon$, find the value of $\delta$ to use.
Let $\epsilon = .04$ to find $\delta$

First set $| (4x + 5) - 13 | < .04$

Based on the absolute value, this means
___ $< 4x - 8 <$ ___ and
___ $< 4x <$ ___ and
___ $< x <$ ___ and

This means we set $\delta = $ (round to four decimal places)

So for any $x$ such that $|x - 2| <$ ___ , we know that $| (4x + 5) - 13 | <$ ___

Please answer all parts of the question.

We wish to show that limx → 2 4x + 5 = 13

If we give an ϵ, find the value of δ to use.
Let ϵ = .04 to find δ

First set | (4x + 5) - 13 | < .04

Based on the absolute value, this means
< 4x - 8 < and
< 4x < and
< x < and

This means we set δ = (round to four decimal places)

So for any x such that |x - 2| < , we know that | (4x + 5) - 13 | <

Please answer all parts of the question.

Added by Anthony S.

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