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Let $L \ne 0$. If $\lim_{t \to x} f(t)=L$, prove that $\lim_{t \to x} \frac{1}{f(t)}=1/L$.

My attempt for the case where $L>0$:

$\large {|\frac{1}{f(t)}-\frac{1}{L}|=\frac{|f(t)-L|}{|L||f(t)|}}$ need an inequality to continue

(1) Using the given, get a bound on $|f(t)-L|$:

Let $\epsilon>0$, by the limit given above, there exists a $\delta>0$ such that:

$0<|t-x|<\delta$ implies $|f(t)-L| < \epsilon$ implies $|f(t)-L| < \frac{1}{2}|L^2|\epsilon$ for a given $\epsilon$, $\delta$ pair.

(2) Getting a bound on $|f(t)|$:

Take $\epsilon=\frac{1}{2}L>0$, then there exists a $\delta_1>0$ such that:

$0<|t-x|<\delta_1$ implies $|f(t)-L| < \frac{1}{2}L$ implies $|f(t)|>\frac{1}{2}L$

(3) Concluding:

For $\delta_{\min}=\min(\delta,\delta_1)>0$ and for $\epsilon>0$

$\large{|\frac{1}{f(t)}-\frac{1}{L}|=\frac{|f(t)-L|}{|L||f(t)|}\leq\frac{2}{|L^2|}}$$ |f(t)-L|<\epsilon$

Is this correct? Also, is this the proper way to approach the problem?
I asked a similar question before btw, but I made a lot of progress since then. Not sure if I had to post this on the old question or what.

Thank you!

Lord_Farin
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wwbb90
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1 Answers1

2

This appears to be correct.

You didn't need to assume $L > 0$, all you needed was $L \neq 0$, or equivalently $\vert L \vert > 0$.

You're reasoning for (2) would need to be modified as follows:

Take $\varepsilon := \frac12 \vert L \vert$, there exists $\delta_1 > 0$ such that if $0 < \vert t - x \vert < \delta_1$ then $\vert f(t) - L \vert < \frac12 \vert L \vert$. Observing the following inequality $$\vert L \vert - \vert f(t) \vert \leq \vert f(t) - L \vert < \frac12 \vert L \vert$$ means that $- \vert f(t) \vert < -\frac12 \vert L \vert$ which means that $\vert f(t) \vert > \frac12 \vert L \vert$, as desired.

Robert Cardona
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