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Suppose $b_n$ is a sequence $>0$ and $b>0$ where $b_n$ converge to $b$. Suppose $z_n=\log b_n$ and $z=\log b$, prove that $z_n$ converge to $z$. I know the definition of limit but not sure how to satisfy the condition

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Do you know that $\log x$ is a continuous function defined on $(0,+\infty)$? What do you know about the erlation of convergent sequences and continuous functions? – Pedro Tamaroff Sep 22 '12 at 16:06
prove by definition definition, we need to find a $N$ from sequence $z_n$ st the condition satisfies by not sure how to find $N$, the only hints is taht we can find such $N$ from sequence $b_n$ – Mathematics Sep 22 '12 at 16:08
In order to be able to help we have to know the rules of the game. What is your definition of $\log$, and what properties of $\log$ (or of sequences, for that matter) are you allowed to use? – Christian Blatter Sep 22 '12 at 18:07

Hint use $ \log \space a - \log\space b = \log \space a/b $ and $\log 1 = 0$

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well, of course i did use but cannot really show the inequality – Mathematics Sep 22 '12 at 15:36
Z - Zn = log b/bn as bn converges to b, zn - z tends to 0 – Ram Sep 22 '12 at 15:39
ya, but from the definition, you gotta show that there should exist an integer $N$ s.t. for all integers $n$, where $n>N, |z_n-z|<\epsilon$ for all $\epsilon >0$ – Mathematics Sep 22 '12 at 15:42
if |zn−z|<ϵ implies |log b/bn| <ϵ => b/bn = e^ϵ (or 10^ϵ depending on log or ln) bn = be^-ϵ – Ram Sep 22 '12 at 15:47
yup, but we have no idea even we got $b_n$ to find the $N$ from sequence {$z_n$} – Mathematics Sep 22 '12 at 15:50


I don't know if the hypothesis that the logarithm is continuous is the best choice, so I'll add something. From any definition of the logarithm, you'll extract that

$$\log x - \log y = \log \frac{x}{y}$$

and that

$$1 - \frac{1}{x} < \log x < x - 1$$

for $x\neq 1$. If $x=1$, we have equalities. From $(2)$, we have that, for $x\neq 1$,

$$\frac{1}{x} < \frac{{\log x}}{{x - 1}} < 1$$

From the squeeze theorem it follows that

$$\mathop {\lim }\limits_{x \to 1} \frac{{\log x - \log 1}}{{x - 1}} = 1$$

from where the logarithm is differentiable at $x=1$, and thus continuous at $x=1$. But the fact that it is continuous at $x=1$ means it is continuous for every $x>0$.

Indeed, pick any sequence $a_n>0$ that converges to $a(>0)$. Then

$$\displaylines{ \mathop {\lim }\limits_{n \to \infty } \log {a_n} = \log a \cr \Leftrightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\log {a_n} - \log a} \right) = 0 \cr \Leftrightarrow \mathop {\lim }\limits_{n \to \infty } \log \frac{{{a_n}}}{a} = 0 \cr} $$

But $\frac {a_n}{a}\to 1$ and $\log 1=0$.

You have that $(b_n)$ is a sequence of positive numbers, that is, $b_n>0\;\forall n$, that converges to $b$. This means that for every $\epsilon>0$ there is an $N_0$ such that, whenever $n\geq N_0$, $|b-b_n|<\epsilon$.

Now, we're setting $z_n=\log \; b_n$. This makes sense for each $n$ for $b_n>0$. Now, we want to prove that, $z_n\to z=\log b$. This means that, for every $\epsilon >0$, there is an $N_1$ such that, whenever $n\geq N_1$, $|z_n-z|<\epsilon$, that is

$$|\log b_n-\log b|<\epsilon$$

But $\log x$ is continuous for $x>0$, this means that for any $\epsilon>0$ there is a $\delta >0$ such that, for all $x$,

$$|x-a|<\delta\implies |\log x-\log b|<\epsilon$$

But then, since $b_n$ converges to $b$, for any $\delta >0$, there will be an $N_\delta$ for which


and consequently

$$|\log b_n-\log b|<\epsilon$$

Thus, we can take $N=N_\delta$. This means that for any $\epsilon >0$, whenever $n\geq N_\delta$ we'll have $$|z-z_n|=|\log b_n-\log b|<\epsilon$$ as desired.

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Setting $$ x_n=\min\{b,b_n\},\ y_n=\max\{b,b_n\}, $$ we have $$ y_n+x_n=b+b_n,\ y_n-x_n=|b-b_n|. $$ Therefore $$ \lim_{n\to \infty}(y_n+x_n)=2b,\ \lim_{n\to \infty}(y_n-x_n)=0, $$ and we deduce that $$ \lim_{n\to \infty}y_n=\lim_{n\to \infty}x_n=b. $$ Since $b>0$, we have $$ \lim_{n \to \infty}\frac{y_n}{x_n}=1. $$ Thus, given $\varepsilon>0$ there is an $N=N(\varepsilon) \in \mathbb{N}$ such that $$ \frac{y_n}{x_n}-1\le \varepsilon \quad \forall\ n\ge N. $$ Hence, for every $n \ge N$ we have $$ |z-z_n|=\left|\int_b^{b_n}\frac{dt}{t}\right|=\int_{x_n}^{y_n}\frac{dt}{t}\le \frac{y_n-x_n}{x_n}=\frac{y_n}{x_n}-1 \le \varepsilon, $$ i.e. $z_n \to z$ as $n \to \infty$.

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