anonymous
  • anonymous
please see attached.
Calculus1
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
The \(\epsilon,\delta\) definition for this type of limit:\[ |x-a|<\delta\implies f(x)<\epsilon \]
anonymous
  • anonymous
The one for \[ \lim_{x\to a}f(x)=0 \]Is \[\forall \epsilon>0\;\exists \delta\;\forall x:\quad |x-a|<\delta \implies |f(x)-0|<\epsilon \]

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anonymous
  • anonymous
The \(\epsilon,\delta\) definition for\[ \lim_{x\to a}\frac{1}{f(x)}=-\infty \]Is given by:\[ \forall \epsilon\;\exists \delta\; \forall x:\quad|x-a|<\delta\implies \frac{1}{f(x)}<\epsilon \]
anonymous
  • anonymous
So: \[ |f(x)-0| = |f(x)| = -f(x) \]Since \(f(x)<0\)
anonymous
  • anonymous
Meaning we know \[ -f(x)<\epsilon_1 \]I'm giving subscripts to distinguish the epsilons.
anonymous
  • anonymous
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: \[ \frac{1}{-f(x)}>\frac 1 {\epsilon_1} \]
anonymous
  • anonymous
Finally we multiply both sides by \(-1\): \[ \frac{1}{f(x)}<-\frac 1{\epsilon_1} \]
anonymous
  • anonymous
Thus if we let \(\epsilon_2 = -1/\epsilon_1\):\[ |x-a|<\delta_2 \implies \frac{1}{f(x)}<\epsilon_2 =-\frac{1}{ \epsilon_1} \]
anonymous
  • anonymous
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.
anonymous
  • anonymous
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!

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