Dear Uncle Colin,

I need to work out the limit of $\frac{2^{3x} - 1}{3^{2x}-1}$ as $x \to 0$, and I don’t have any ideas at all. Do you?

- Fractions Rotten, Exponents Generally Excellent

Hi, FREGE, and thanks for your message!

There are a couple of ways to approach this: you should choose between them depending on whether you’re allowed to use L’Hopital’s rule.

L’Hopital’s rule

You can use L’Hopital’s rule ((discovered by Bernoulli)) whenever you’re working with the limit of a fraction, such that both the top and the bottom evaluate to zero in the limit; or if they both tend to infinity in the limit. When this condition is met, the limit of the fraction $\frac{f(x)}{g(x)}$ is the same as the limit of the derivatives of the two bits: $\frac{f’(x)}{g’(x)}$.

So, in this case, $f(x) = 2^{3x}-1$, which is zero when $x=0$, and differentiates to $f’(x) = 3\ln(2) 2^{3x}$. When $x=0$, that’s just $3\ln(2)$.

The bottom goes very similarly: $g(x) = 3^{2x} - 1$ (which is zero in the limit) and $g’(x) = 2\ln(3) 3^{2x}$, which is $2\ln(3)$.

That makes the limit of the original fraction $\frac{3\ln(2)}{2\ln(3)}$.

The “proper” way

Alternatively, we can rewrite the fraction as $F(x) = \frac{e^{3\ln(2)x} - 1}{e^{2\ln(3)x}-1}$.

I know some things about $e^{kx}-1$, let me tell you. For example, I know that $kx < e^{kx}-1 < k\left(x+x^2\right)$, certainly for small $0<x<1$.

That’s helpful! It means that as long as $0 < x < 1$, we know $\frac{3\ln(2) x}{2\ln(3) (x + x^2)} < F(x) < \frac{3\ln(2) (x + x^2)}{2\ln(3)x}$. (Note that the first fraction has the smallest possible top and largest possible bottom, and vice versa).

Now let’s simplify:

$\frac{3\ln(2)}{2\ln(3) (1+x)} < F(x) < \frac{3\ln(2) ( 1+x)}{2\ln(3)}$.

As $x$ approaches 0, the lower fraction approaches $\frac{3\ln(2)}{2\ln(3)}$ - and so does the upper one! Since $F(x)$ is between those, it must also approach the same limit ((I never know whether to call this the ‘squeeze theorem’ or the ‘sandwich theorem’.)).

I hope that helps!

- Uncle Colin