How to answer the question from Calculus by Michael Spivak Chapter 5 problem 34

Question

I want to prove that $$\lim_{x\to 0+} f(\frac{1}{x}) = \lim_{x\to ∞}f(x)$$

I know the definition of the left side that is:

• ∀ ε > 0 ∃ δ > 0: if 0 < x < δ then |$f(\frac{1}{x})$ -L| <ε

And I want to prove that is the same as:

• ∀ ε > 0 ∃ N > 0: if x > N then |f(x)-L| <ε

I am a beginner so I apologize if thi is too basic.

Answer 1

You can take $N$ as you stated, then take $0<\delta<\frac{1}{N}$, then $\frac{1}{\delta}>N$ so if $|x|<\delta$, you have $|\frac{1}{x}|>\frac{1}{\delta}>N$.

And you need to consider the situation that $f$ diverges.

Answer 2

Note that if $$ 0<|x|<\epsilon\implies \left|\frac{1}{x} \right|>\frac{1}{\epsilon} $$ You may make the quantity on the far left as large as you want by taking $\epsilon$ small.

Answer 3

I know I'm a little late in answering this question but as I'm solving Spivak Calculus now and came across this question, I thought to answer it.

Now, as you have rightly written in the definition, let $\lim_{x\to ∞}f(x)=L$ then, for every $\epsilon \gt 0$, there is $x$ such that for all $x \gt N \implies |f(x) - L| \lt \epsilon$

Let $\frac{1}{\delta} = N$

Now, $0 \lt x \lt \delta \implies {\frac{1}{x}} \gt \frac{1}{\delta} \implies \frac{1}{x} \gt N \implies |f(\frac{1}{x}) - L| \lt \epsilon $

Hence $0 \lt x \lt \delta \implies |f(\frac{1}{x}) - L| \lt \epsilon $ which means $\lim_{x\to 0^+}f(x)=L$