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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

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

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I think, we have no any problems: $$\lim_{x\rightarrow+\infty}\frac{(x-1)e^x}{x^2}=\lim_{x\rightarrow+\infty}\frac{(x-1)e^x+e^x}{2x}=\lim_{x\rightarrow+\infty}\frac{xe^x+e^x}{2}=+\infty.$$

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