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my exercise is to proof the following: if $(x_n)_{n\in\mathbb N}$ is a positive sequence, and if $$\lim_{n\rightarrow+\infty} x_n =x,\,\, \mbox {where } x>0$$ then $$\lim_{n\rightarrow+\infty} \log x_n=\log x$$ I have no idea how that works and I would be happy if someone could help me.

EDIT: I am not allowed to use the information that the logarithm is a countinous function.

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    The logarithm is a continuous function ... – Martin R Jul 03 '19 at 14:08
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    Yes I knew that already, but i don't really how that helps. Can someone give me a hint? –  Jul 03 '19 at 14:13
  • Write down the definition of continuity at the point $x$. It is exactly what you need. – Martin R Jul 03 '19 at 14:14
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    @MartinR: that depends on which definition of continuity you use. – TonyK Jul 03 '19 at 14:15
  • What did you try? Did you look through the relevant definitions one more time? If not that might bring some clarity. –  Jul 03 '19 at 14:16
  • So far we have not defined what a continous function is. So I am not allowed to use this. Are there any other ideas? –  Jul 03 '19 at 14:20

2 Answers2

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I think that your question might be a duplicate or similar to this one

Logarithm is a continuous function, and in general, if $\lim\limits_{x \to c} g(x) = b$, and $f$ is continuous at $b$, then $f\left(\lim\limits_{x \to c} g(x)\right) = f(b) = \lim\limits_{x \to c} f\big(g(x)\big)$. – Henry Swanson

Arhiliuc Cristina
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  • I am not allowed to use the information that the logarithm is a countinous function. Without this restriction, the exercise would be easy. But thanks for answering! –  Jul 03 '19 at 14:21
  • Please add it to the initial condition, so that the others would try to do it differently – Arhiliuc Cristina Jul 03 '19 at 14:24
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Well, think about

$$| \log x_n -\log x| = \left|\log \frac{x_n}{x} \right| < \epsilon.$$

It would be implied by $e^{-\epsilon} <\frac{x_n}{x} < e^\epsilon.$

Context would help here. If you already have a theorem about the quotient of limits, then you're about done.

B. Goddard
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