Compute the limit of exponentials involving 4^x

Expert intro: This limit is a good place to factor carefully before reaching for l’Hôpital. A small rewrite turns the numerator into something much friendlier.

Detailed walkthrough

We need

$\lim_{x\to 0}\frac{4^{5x}-4^{4x}}{3(4^x-1)}.$

Step 1: Factor the numerator

Write

$4^{5x}-4^{4x}=4^{4x}(4^x-1).$

Then the expression becomes

$\frac{4^{4x}(4^x-1)}{3(4^x-1)}.$

For $x\neq 0$, cancel $4^x-1$:

$\frac{4^{4x}}{3}.$

Step 2: Take the limit

Now let $x\to 0$:

$4^{4x}\to 4^0=1.$

So the limit is

$\boxed{\frac{1}{3}}.$

💡 Pitfall guide

Do not cancel $4^x-1$ before factoring correctly; the factor is present in the numerator only after the rewrite. Also, there is no need for a derivative formula here. The algebra already resolves the indeterminate form.

🔄 Real-world variant

If the denominator were $k(4^x-1)$ instead of $3(4^x-1)$, the same method would give

$\lim_{x\to 0}\frac{4^{5x}-4^{4x}}{k(4^x-1)}=\frac{1}{k}.$

The constant in front just scales the final answer.

🔍 Related terms

exponential limits, factoring, indeterminate form