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So I have to prove that:

$$\lim_{x\to a} f(x) = +\infty \implies \lim_{x\to a} \frac{1}{f(x)} = 0$$

By the pricese definition of a limit, if $\lim_{x\to a} f(x) = +\infty$, then:

$$\forall M>0, \exists \delta_1 | 0<|x-a|<\delta_1 \implies f(x)>M \tag{1}$$

And I have to prove that is always leads to $\frac{1}{f(x)} < \epsilon$, because:

$$\forall \epsilon>0, \exists \delta_2 | 0<|x-a|<\delta_2 \implies \left|\frac{1}{f(x)} - 0\right|<\epsilon $$

So, using $(1)$, I can always choose $M = \frac{1}{\epsilon}$, so:

$$\forall M>0, \exists \delta_1 | 0<|x-a|<\delta_1 \implies f(x)>M \implies \frac{1}{f(x)} < \frac{1}{M} = \frac{1}{\frac{1}{\epsilon}} = \epsilon$$

So $\frac{1}{f(x)} < \epsilon$

Is my proof correct?

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  • $\begingroup$ Pick $\delta$ carefully. $\endgroup$ – IAmNoOne Apr 24 '14 at 3:37
  • $\begingroup$ @Nameless something's wrong? Could you give me a hint? $\endgroup$ – Marter Js Apr 24 '14 at 3:41
  • $\begingroup$ @MarterJs I don't see any error. It looks good to me. I have put a suggestion on how to streamline the proof into an answer below, but your logic seems correct. $\endgroup$ – Ryan Sullivant Apr 24 '14 at 3:48
  • $\begingroup$ You cannot choose $M$. You must choose a $\delta$ such that it works for any $M>0$. $\endgroup$ – chubakueno Apr 24 '14 at 3:52
  • $\begingroup$ @chubakueno Even though the o.p. says "choose $M$" he's really just saying there is a $\delta$ (which exists since the limit of $f(x)$ is $\infty$) for this specific $M$. $\endgroup$ – Ryan Sullivant Apr 25 '14 at 8:20
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It looks good, but you could clean it up a bit. For example, you don't need the "$\forall M>0$" on the third to last line. The idea is completely correct thought.

To show that $\lim_{x \to a} \frac{1}{f(x)} = 0$ given $\varepsilon >0$ we need to find $\delta>0$ such that if $0< |x-a|< \delta$ then $\left |\frac{1}{f(x)} \right| < \varepsilon$.

Now, let $\varepsilon_0 >0$ be given. As you say, since $\lim_{x\to a} f(x) = \infty$ there is a $\delta_0 >0$ such that if $0<|x-a| < \delta_0$ then $f(x) > M_0$ where $M_0 = \frac{1}{\varepsilon_0}$.

Since $f(x) > M_0$, for $0<|x-a|<\delta_0$ we get $$\left|\frac{1}{f(x)} \right | < \frac{1}{M_0} = \varepsilon_0$$

Hence, $\lim_{x\to a} \frac{1}{f(x)} = 0$.

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