If function $f$ defined in an pocked neighbourhood of $x_0$, $N*\lambda(x_0)$ and positive there (meaning $f(x)>0$ to all $x\in N*\lambda(x_0)$).

If : $$\lim_{x \to x_0} f(x)=0$$


$$\lim_{x \to x_0} \frac{1}{f(x)}=\infty$$

We need to proof that (The proof as shown in the book):

All $N>0$ exist $\delta>0$ so all $x\in N*\delta(x_0)$ appiles $f(x)>N$.

Exist $N>0$.

Since $\lim_{x \to x_0} f(x)=0$ we know that $0<\delta<\lambda$ so all $x\in N*\delta(x_0)$ appiles $0<f(x)<\frac{1}{N}$ Therefore in this neighbourhood $\frac{1}{f(x)}>N$

I don't understand how they conclude that $f(x)>\frac{1}{N}$

  • $\begingroup$ I believe you mean that $\lim 1/f(x) = \infty$. $\endgroup$ – Joel Aug 13 '14 at 13:53
  • $\begingroup$ @Joel Indeed, Edited. Thanks $\endgroup$ – JaVaPG Aug 13 '14 at 14:02
  • $\begingroup$ Where do they conclude $f(x) > \frac1N$? $\endgroup$ – Frunobulax Aug 13 '14 at 14:05

Since $f(x) \to 0$ as $x \to x_0$ we know that for any $0 < \epsilon$ there is a $\delta$ for which $| f(x) - 0 | < \epsilon$ for all $x$ such that $|x-x_0| < \delta$. This is the definition of a limit.

Here we choose our $\epsilon$ to be $1/N$ for a natural number $N$. Then a corresponding $\delta$ was chosen so that $$0 < |f(x)-0| = |f(x)| = f(x) < 1/N$$ for all $x \in N_\delta(x_0)$. (Note that we could drop the absolute value bars, since $f(x)$ is positive)

From here we can multiply both sides of $0 < f(x) < 1/N$ by $N$ and divide both sides by $f(x)$, since $f(x)$ is positive, to find $$0 < N < 1/f(x).$$ Hence as $x$ gets closer to $x_0$, $1/f(x)$ gets arbitrarily large.


I think there's a typo: You say "applies $f(x)>N$", but you want actually $\frac{1}{f(x)}>N$.

The reasoning is just in the second last line: Once you know


you conclude by taking the whole inequality $f(x)<\frac{1}{N}$ to the power -1. And all the quantities are positive so the statement follows. Or does the difficulty arise before that line?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.