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

MadViolinist

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!

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
$$\lim_{x \rightarrow \infty} \ln(u(x)) = \ln(L)$$.

MadViolinist

Thanks for the answer!

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?

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##.

MadViolinist

Thanks again!