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The following is an answer to the proof of

$$ \lim \limits_{x\to 0^+}f\left( \frac{1}{x} \right)=\lim \limits_{x\to \infty}f(x)$$

If $l=\lim \limits_{x\to \infty}f(x)$, then for every $\epsilon>0$ there is some $N$ such that $|f(x)-l|<\epsilon$ for $x>N$, and we can clearly assume that $N>0$. Now if $0<x<\frac{1}{N}$, then $\frac{1}{x} > N$, so $|f\left(\frac{1}{x}\right)-l|<\epsilon$. Thus, $\lim \limits_{x\to 0^+}f\left( \frac{1}{x} \right)=l$.

I don't get how if $\frac{1}{x} > N$, we can conclude that $|f\left(\frac{1}{x}\right)-l|<\epsilon$.

Git Gud
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mauna
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1 Answers1

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Let $L=\lim \limits_{y\to \infty}f(y)$. Then by definition, we know that:

For every $\epsilon^*>0$, there is some $N>0$ such that $|f(y)-L|<\epsilon^*$ for all $y>N$.

We want to show that $\lim \limits_{x\to 0^+}f\left( \frac{1}{x} \right)=L$. That is, we want to show that:

For every $\epsilon>0$, there is some $\delta>0$ such that if $0<x<\delta$, then $|f(1/x)-L|<\epsilon$.

With that in mind, choose any $\epsilon>0$. Now let $\epsilon^*=\epsilon$. Using the first definition, let $\delta=\dfrac{1}{N}$, where $N$ corresponds to our selection of $\epsilon^*=\epsilon$.

Now suppose that $0<x<\delta=\dfrac{1}{N}$. Then we know that $\dfrac{1}{x}>N$. Now let $y=\dfrac{1}{x}$. Since $y=\dfrac{1}{x}>N$, it follows by the first definition that $\left|f\left(\dfrac{1}{x}\right)-L\right| < \epsilon^*=\epsilon$, as desired.

Adriano
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