# Proving continuity of Logarithm using $\delta$-$\epsilon$ [duplicate]

Say we wanted to prove the continuity of the logarithm using the $$\delta - \epsilon$$ proof (and using the definition of the log as the inverse of the exponential). For any log base $$a>1$$ Starting with $$|\log_a({x_1}) - \log_a({x_2})|<\epsilon$$, We can find that if $$\frac{x_1}{x_2} < a^\epsilon$$ (and $$\frac{x_1}{x_2}), then $$|\log_a({x_1}) - \log_a({x_2})|<\epsilon$$ as desired. But for the $$\delta$$ part to come in, we need it in the form of $$|x_1 - x_2|$$, not $$\frac{x_1}{x_2}$$.

So then I thought that maybe if $$|x_1 - x_2| < \frac{x_1}{x_2} -1 < a^\epsilon -1$$ then we would have $$|\log_a({x_1}) - \log_a({x_2})|<\epsilon$$ which would mean that the function is continuous at $$x_1$$ and since it was arbitrary it is continuous at all x (right?).

The reason I think this is true is because we want the distance between $$x_1$$ and $$x_2$$ to be very small, so $$a^\epsilon -1$$ (which we would call $$\delta$$) would get really small as $$\epsilon$$ gets small. Is there a better way to arrive at this value of $$\delta$$? Assuming it is a valid answer

• What definition of $\log$ are you using? One way this can be done is to define $\log$ as $\log(x)=\int_1^x\frac1t\mathrm{d}t$, and then use continuity of Riemann integrals. Generally the continuity of the logarithmic function is not proved with a standard $\delta/\epsilon$ proof – Václav Mordvinov Oct 11 '18 at 20:10
• We are using the inverse of the exponential definition. We haven't covered any other methods in my class so far so I'm hoping to answer it with $\delta - \epsilon$ – Riley H Oct 11 '18 at 20:15
• @RileyH How are you defining the exponential function? – Jack M Oct 11 '18 at 20:16

What you propose is the following proof. In what follows I'll assume $$a>1$$, I can't be bothered to worry about whether or not the details apply if $$a\leq 1$$.

Let $$\epsilon>0$$ and $$x_0>0$$. We want to find a small enough $$\delta$$ such that for $$|x-x_0|<\delta$$, $$|\log_a(x)-\log_a(x_0)|<\epsilon$$. We have

$$|\log_a(x)-\log_a(x_0)|=|\log_a(\frac x{x_0})|$$

and you suggest that if we can guarantee $$\frac x{x_0}, then we're done. First of all, this is completely untrue as stated because that could mean $$\frac x{x_0}$$ could be close to $$0$$ in which case that logarithm would be massively negative, not what we want. What we want is for $$\frac x{x_0}$$ to be close to $$1$$. A better condition would be $$a^{-\epsilon}<\frac x{x_0}. In that case, we would have, by the strictly increasing nature of the $$\log$$ function:

$$-\epsilon<\log_a(\frac x{x_0})<\epsilon$$ $$|\log_a(\frac x{x_0})|<|\epsilon|$$

As you say, we need to show that we can guarantee $$a^{-\epsilon}<\frac x{x_0} for $$|x-x_0|<\delta$$ by choosing a small enough $$\delta$$. Probably the cleanest way to do this is to separate the cases $$x>x_0$$ and $$x. For $$x>x_0$$ you just need $$\frac x{x_0} since the other inequality is trivial, letting $$\delta=a^\epsilon-1$$ we get immediately $$|\frac x{x_0}-1|<\delta$$, the case $$x is similar. So that part of the proof isn't that mysterious.

The mysterious part is figuring out if this is really a proof. What facts about logarithms and exponentials did we implicitly use here? Can we justify all of them? We used the following:

1. $$\log_a$$ is a real function defined for all $$x>0$$.
2. The formula $$\log(x)-\log(y)=\log\frac{x}{y}$$.
3. $$\log$$ is strictly increasing.
4. $$a^x$$ is strictly increasing (this is equivalent to (3) under your definitions).
5. $$a^0=1$$.

I was originally going to write an answer about how we would first have to prove that $$a^x$$ is continuous, but I don't think I've used that assumption anywhere. I suppose the key is in assumption (2), which is such a powerful assumption (it basically uniquely characterizes the logarithm) that it implicitly gives us the continuity of $$\log$$ automatically. In fact, I think it should be possible to skip any mention of $$a^x$$ and show $$\log$$ is continuous just using the assumption $$\log(x)+\log(y)=\log(xy)$$.