N's the Limit

Here's some background info for my solution: Lambert $w$ function

In this problem, you can simply use inspection to guess the value of $n$ . Here is a more rigorous and generalised approach:

I just used l'Hôpital's rule; $\displaystyle \frac{d}{dx} (x^n - 3^n) = 108$

Evaluating results in:

$\displaystyle n(3^{n-1}) = 108$

I am giving a more general way of solving the equation above:

Leaving the expression in Lambert form:

$\displaystyle ne^{\ln \left(3\right)\left(n-1\right)}=108$

Using the substitutions:

$\displaystyle u = n\ln \left(3\right)$

$\displaystyle n=\frac{u}{\ln \left(3\right)}$

Simplifying yields $\displaystyle \frac{e^{u-\ln \left(3\right)}u}{\ln \left(3\right)}=108$

Which in Lambert form is:

$\displaystyle e^uu=324\ln \left(3\right)$

So $\displaystyle u=4\ln \left(3\right)$ .

Substituting back gives $\boxed{n = 4}$

Given that

$\begin{aligned} \lim_{x \to 3} \frac {x^n - 3^n}{x-3} & = 108 & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies}} \\ \lim_{x \to 3} \frac {nx^{n-1}}1 & = 108 & \small \blue{\text{Differentiate up and down w.r.t. }x} \\ n3^{n-1} & = 108 \\ \implies n & = \boxed 4 \end{aligned}$

---

Reference: L'Hôpital's rule

Evaluating the LHS, we get :

$\begin{aligned} \lim_{x \to 3} \dfrac{x^{n} - 3^{n}}{x-3} &= n \times 3^{n-1} \\ n \times 3^{n-1} &= 108 \\ n \times 3^{n-1} &= 4 \times 27 \\ &\boxed{n = 4} \end{aligned}$

$(4 \times 3^{4-1} = 4 \times 3^{3} = 4 \times 27 = 108)$

The form of the limit is exactly the derivative of $x^n$ at the point $x=3$ - that is $n\cdot 3^{n-1}=108$

Since we know we're after an integer, we can solve by inspection and find $n=\boxed4$ .