Q.Find '$n$', if $\lim_{x \to 2} \dfrac{x^n - 2^n}{x - 2} = 80$, $n \in \mathbf{N}$.

Concept: Limit of a polynomial — this is the standard form of the derivative of $x^n$ at $x = 2$.

Recall the limit definition of the derivative:

Here $a = 2$, so:

We are given this equals $80$:

Now test natural numbers $n$: …

The expression $\frac{x^n - 2^n}{x - 2}$ looks familiar — it’s the difference quotient for the function $f(x) = x^n$ at the point $x = 2$. As $x \to 2$, this quotient approaches the derivative $f'(2)$.

So the problem is really asking: for which natural number $n$ does $f'(2) = 80$?

1. Recognize the limit as a derivative. For any function $f$,

Here $f(x) = x^n$ and $a = 2$, so

2. Compute the derivative. The power rule gives $\frac{d}{dx}(x^n) = n x^{n-1}$. Evaluating at $x = 2$:

3. Set equal to 80. We have

4. Solve for natural $n$. Try small $n$:
   • $n=1$: $1 \cdot 2^{0} = 1$
   • $n=2$: $2 \cdot 2^{1} = 4$
   • $n=3$: $3 \cdot 2^{2} = 12$
   • $n=4$: $4 \cdot 2^{3} = 32$
   • $n=5$: $5 \cdot 2^{4} = 5 \cdot 16 = 80$ — works.
   • $n=6$: $6 \cdot 2^{5} = 6 \cdot 32 = 192$, too large. …