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Lunna [17]
3 years ago
7

Pls help. i never understood this rule.

Mathematics
1 answer:
Dimas [21]3 years ago
8 0
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
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