Question

Pls help. i never understood this rule.

Answer 1

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


`\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} \\\\\\ \textit{so we can use L'Hopital by instead using the derivatives} \\\\\\ \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`