Question

Check for work

Answer 1

Given:

\(f(x)=5x-2\)
\(c=3, L=13, \epsilon = 1.75\)

\(0<|x-c|<\delta \rightarrow |f(x)-L|<\epsilon\)

Answer 2

My work:

\(|f(x)-L|<\epsilon\)
\(|(5x-2)-13|<\epsilon\)
\(x<\dfrac{\epsilon}{5}\)


Then
\(\delta<\dfrac{\epsilon}{5}\)
\(\delta<\dfrac{1.75}{5}\)
\(\delta<0.35\)

Answer 3

@AZ c;

Answer 4

Whoops, meant

\(x-3<\dfrac{\epsilon}{5}\)

Answer 5

Ah, yes, the Epsilon-Delta definition of a limit. Fun stuff /s

\[ \lim_{x \rightarrow a} f(x) = L\]
If
\[\forall~\epsilon>0 ~~\exists~\delta>0 \]
Then
\[0<|x-a|<\delta \implies |f(x)-L|<\epsilon \]

Answer 6

\(\color{#0cbb34}{\text{Originally Posted by}}\) @dude
Whoops, meant

\(x-3<\dfrac{\epsilon}{5}\)
\(\color{#0cbb34}{\text{End of Quote}}\)

Yes yes

Good job

Answer 7

Thanks :D

Answer 8

You're welcome :D