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