Dominating Matter

the correct solution without use L'hopital Rule.

$\large \lim_{x\to\infty} (4^{x} + 1)^{\frac{1}{x}} = \large e^{lim_{x\to\infty} \frac{1}{x} \cdot ln (4^{x} +1)} = e^{lim_{x\to\infty} \frac{1}{x} \cdot ln(4^{x})} =$ $= \large e^{lim_{x\to\infty} \frac{x}{x} \cdot ln 4} = \large e^{ln 4} = \large \boxed{4}$

As $x$ gets large, $4^x \approx 4^x + 1$ when you compare their magnitude. Since the inverse power primarily deals with the property of magnitude, the limit approximates to be $(4^x)^\frac{1}{x}$ which is equal to 4.