# Can L'Hospital's rule be applied to a limit of the form of (inf * inf)/inf?

My teacher wanted us to resolve this limit using L'Hospital. $$\lim_{x\to +∞} \frac{(e^x)(x-1)}{x^2}$$

But I can't understand why she derived the numerator, as it's an indefinite form (inf * inf) and I don't understand why L'Hospital's rule can be directly applied. Thank you in advance

• Sure because \infty*\infty=\infty and you must verify the conditions of de l’Hopital hypothesis – Federico Fallucca Feb 18 '19 at 8:42
• The denominator is $x\cdot x$, and is therefore also of the form $\infty\cdot\infty$, yet that doesn't seem to bother you. Why not? – Arthur Feb 18 '19 at 8:43
• It is in the form $\frac {\infty} {\infty}$ and L'Hopital Rule is applicable. – Kavi Rama Murthy Feb 18 '19 at 8:43
• It is not very well-known that L'Hospital's Rule applies to the form "$\text{anything} /\infty$" and hence it is applicable here. – Paramanand Singh Feb 18 '19 at 15:47

If you expand the numerator as $$xe^x-e^x$$, then the expression at the limit is of the form $$\infty/\infty$$ and L'hopital's rule applies.
I think, we have no any problems: $$\lim_{x\rightarrow+\infty}\frac{(x-1)e^x}{x^2}=\lim_{x\rightarrow+\infty}\frac{(x-1)e^x+e^x}{2x}=\lim_{x\rightarrow+\infty}\frac{xe^x+e^x}{2}=+\infty.$$