# How would I prove that ln (lim (u)) = lim (ln (u))

1. Feb 12, 2013

We were working with L'hospital's rule and my teacher said that the teacher before him told him that this was true:

lim (x→∞) [ln u] = ln ( lim (x→∞) u), where u is a continuous function.

My teacher has never found a proof for this, although it works every time. Does anyone know how to prove this? Thanks!

2. Feb 12, 2013

### jbunniii

I guess you are implicitly assuming that $\lim_{x \rightarrow \infty}u(x)$ exists, because otherwise the statement doesn't make any sense. So if we put $L = \lim_{x \rightarrow \infty}u(x)$ and $x_n$ is any sequence such that $x_n \rightarrow \infty$ as $n \rightarrow \infty$, we have a corresponding sequence $u_n = u(x_n)$ such that $u_n \rightarrow L$ as $n \rightarrow \infty$. We may now write
$$\ln(L) = \ln(\lim_{x \rightarrow \infty}u(x)) = \ln(\lim_{n \rightarrow \infty} u_n) = \lim_{n \rightarrow \infty} \ln(u_n) = \lim_{n \rightarrow \infty} \ln(u(x_n))$$
The third equality holds because $\ln$ is continuous at $L$. Thus we have established that
$$\lim_{n \rightarrow \infty} \ln(u(x_n)) = \ln(L)$$
This is true for any sequence $x_n \rightarrow \infty$, so we may conclude that
$$\lim_{x \rightarrow \infty} \ln(u(x)) = \ln(L)$$.

3. Feb 12, 2013

However, I do not get the step where ln(limn→∞ un)=limn→∞ ( ln(un)). How did you change the order of the natural log and the limit?

4. Feb 12, 2013

### jbunniii

If $f$ is any function that is continuous at $L$, and $\lim_{n \rightarrow \infty} u_n = L$, then $\lim_{n \rightarrow \infty} f(u_n) = f(\lim_{n \rightarrow \infty} u_n) = f(L)$. You can prove this quite easily using the epsilon-delta definition of continuity.

Assuming that theorem, all you need is the fact that $\ln$ is continuous at $L$, where $L$ is any real positive number. (By the way, positivity of $L$ is another assumption that needs to be added to the problem statement, otherwise the equation makes no sense.)

How to prove that $\ln$ is continuous depends on how you defined $\ln$. One standard definition is
$$\ln(x) = \int_{1}^{x} \frac{1}{t} dt$$
If we use that definition, then
\begin{align} |\ln(x+h) - \ln(x)| &= \left|\int_{1}^{x+h} \frac{1}{t} dt - \int_{1}^{x} \frac{1}{t}\right|\\ &= \left|\int_{x}^{x+h}\frac{1}{t} dt\right|\\ \end{align}
If $h > 0$ then we have the bound
$$\left|\int_{x}^{x+h}\frac{1}{t} dt\right| \leq \left|\int_{x}^{x+h}\frac{1}{x} dt\right| = \left| \frac{h}{x} \right|$$
which we can make as small as we like as $h \rightarrow 0$. A similar argument holds for $h < 0$.

5. Feb 12, 2013