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

Then:

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

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 $f(x)>\frac{1}{N}$

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  • $\begingroup$ 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
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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.

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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

$0<f(x)<\frac{1}{N}$

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?

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