L'Hopital rule only applies to indeterminate cases of 0/0 or ∞/∞ or 0*0 or ∞ * ∞.
and you'd need to make the expression a rational, where the numerator and denominator gives you 0/0 or ∞/∞, then you apply L'Hopital.
the only candidate there is
[tex]\bf \lim\limits_{x\to \infty}\cfrac{2x^2-1}{3x+1}\implies \lim\limits_{x\to \infty}\cfrac{2(\infty)^2-1}{3(\infty)+1}\implies \cfrac{\infty}{\infty}
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\textit{so we can use L'Hopital by instead using the derivatives}
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\lim\limits_{x\to \infty}\cfrac{2x^2-1}{3x+1}\stackrel{LH}{\implies }\lim\limits_{x\to \infty}\cfrac{4x}{3}\implies \cfrac{4(\infty)}{3}\implies \infty[/tex]
