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Spivak chapter $5$, question $34.\:\:$

Prove that $$ \displaystyle \lim_{x \to 0+} f\left(\dfrac{1}{x}\right) = \lim _{x \to \infty} f(x)$$

I have no idea on how to even start the proof, so any help would be much appreciated. Thanks

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Let us call $\lim\limits_{x \to 0^+}f(\frac 1 x) = L$

Consider $ |f(x) - L| $ and substitute $t = \frac 1 x$ for $x \gt 0$. Given any $\epsilon \gt 0$ there is a $\delta \gt 0$ such that $f|(\frac 1 x) - L| \lt \epsilon$ whenever $0 \lt \frac 1x \lt \delta$. That is $|f(x) - L| \lt \epsilon $ whenever $ \frac 1 \delta \gt x$. That is given any $\epsilon \gt 0$ there is a positive quantity $M(= \frac 1 \delta)$ such that $x \gt M \implies |f(x) - L| \lt \epsilon \implies \lim\limits_{x \to \infty} f(x) = L$.

$\mathscr {Q.E.D.}$

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Let us try not to suppose that any of the limits exist.

$\lim_{0^+} f(1/x) =L$ is equivalent to:

$$ \forall r>0\ \ \exists \delta>0\ \ 0<x<\delta\Rightarrow -r < f(1/x) - L < r\\ \forall r>0\ \ \exists N>0\ \ 0<x<\frac 1N\Rightarrow -r < f(1/x) - L < r\\ \forall r>0\ \ \exists N>0\ \ 1/x > N\Rightarrow -r < f(1/x) - L < r\\ \forall r>0\ \ \exists N>0\ \ s > N\Rightarrow -r < f(s) - L < r\\ $$ which is equivalent to $$\lim_{\infty} f =L$$

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