Question

please see attached.

Answer 1

The \(\epsilon,\delta\) definition for this type of limit:\[
|x-a|<\delta\implies f(x)<\epsilon
\]

Answer 2

The one for \[
\lim_{x\to a}f(x)=0
\]Is \[\forall \epsilon>0\;\exists \delta\;\forall x:\quad
|x-a|<\delta \implies |f(x)-0|<\epsilon
\]

Answer 3

The \(\epsilon,\delta\) definition for\[
\lim_{x\to a}\frac{1}{f(x)}=-\infty
\]Is given by:\[
\forall \epsilon\;\exists \delta\; \forall x:\quad|x-a|<\delta\implies \frac{1}{f(x)}<\epsilon
\]

Answer 4

So: \[
|f(x)-0| = |f(x)| = -f(x)
\]Since \(f(x)<0\)

Answer 5

Meaning we know \[
-f(x)<\epsilon_1
\]I'm giving subscripts to distinguish the epsilons.

Answer 6

It's a limit so let's only worry about arbitrarily small \(\epsilon_1\).\[
0<-f(x)<\epsilon_1<1
\]When we take the inverse of both sides, the inequality flips: \[
\frac{1}{-f(x)}>\frac 1 {\epsilon_1}
\]

Answer 7

Finally we multiply both sides by \(-1\): \[
\frac{1}{f(x)}<-\frac 1{\epsilon_1}
\]

Answer 8

Thus if we let \(\epsilon_2 = -1/\epsilon_1\):\[
|x-a|<\delta_2 \implies \frac{1}{f(x)}<\epsilon_2 =-\frac{1}{ \epsilon_1}
\]

Answer 9

Proving: \[
\lim_{x\to a}f(x)=0 \implies \lim_{x\to a}\frac{1}{f(x)} = -\infty
\]
1) They give us an \(\epsilon_2\)
2) We convert it to \(\epsilon_1=01/\epsilon_2\)
3) Since we know our limit exists, we know \(\delta \) exists.
4) Our previous steps ensured \(\delta\) will work for both limits.

Proving the converse is the same thing but saying it backwards.

Answer 10

Good ans. However, it is only considered partially proved.
For if and only if statement. You have only proved the "only if" statement.
Which means limx→af(x)=0⟹limx→a1f(x)=−∞
However, the reverse needs to be proved as well. "if" statement

Thanks for the help!