let's check these folks using LH, or L'Hopital rule, since they're indeterminate types.
[tex]\bf \textit{2nd pair listed gives }\\\\
\lim\limits_{x\to \infty}\ \cfrac{x^2}{e^{4x}}\implies \underline{LH}\quad \lim\limits_{x\to \infty}\ \cfrac{2x}{4e^{4x}}[/tex]
on this pair, the denominator is moving faster than the numerator, and thus, yielding a much larger denominator on every iteration, moving towards a limit of 0.
[tex]\bf \textit{3rd pair listed gives }\\\\
\lim\limits_{x\to \infty}\ \cfrac{[ln(x)]^3}{x}\implies \underline{LH}\ \lim\limits_{x\to \infty}\ \cfrac{3ln(x)\cfrac{1}{x}}{1}\implies \lim\limits_{x\to \infty}\ \cfrac{3ln(x)}{x}[/tex]
on this pair, the denominator, again, is moving faster, since ln() results in an exponent for the constant "e", and thus is a much smaller then the denominator on every iteration, thus, the denominator gets much larger and moving towards 0 as limit as well.
[tex]\bf \textit{4th pair listed gives }\\\\
\lim\limits_{x\to \infty}\ \cfrac{\sqrt{x}}{e^x}\implies \underline{LH}\lim\limits_{x\to \infty}\ \cfrac{\frac{1}{2\sqrt{x}}}{e^x}\implies \lim\limits_{x\to \infty}\ \cfrac{1}{2e^x\sqrt{x}}[/tex]
on this pair, the numerator is just a constant, thus is static, and the denominator is moving onwards on every iteration, thus moving also again towards 0.
now, let's take a peek of the 1st pair for f(x) and g(x)
[tex]\bf \textit{1st pair listed gives }\\\\
\lim\limits_{x\to \infty}\ \cfrac{10+e^{-x}}{5x}\implies \underline{LH}\ \lim\limits_{x\to \infty}\ \cfrac{10-\frac{1}{e^x}}{5}
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\textit{notice, the limit of each term, specifically of }\frac{1}{e^x}\textit{ goes to \underline{0}}
\\\\\\
\textit{so the function turns to }\lim\limits_{x\to \infty}\ \cfrac{10-0}{5}\implies \boxed{2}[/tex]
