Limits with indeterminant form, using L'Hopital's rule, help?

Question

I'm a bit confused on why the answer is what it is:

lim x -> 0+ of (e^(x) - x)/(x^3)

In words, the limit as x goes to zero from the right of e^x minus x all divided by x^3

The answer turns out to be infinity (diverges, same thing). Why is that?

I get e^x/6x and then I'm stuck, since you can't continue on with l'hopital's rule because you get 1/0

Any explanations would be awesome. Thanks geniuses. :D

? · Accepted Answer

You can't use L'Hopital in this case because it is not in the required form of 0/0 or ∞/∞. See:

(e^x - x) / x^3
(e^0 - 0) / 0^3
(1 - 0) / 0
1 / 0

So, in this case, the limit is +∞.

anatoliy29 · Answer

You keep differentiating the top and the bottom until it's not infinity/infinity anymore

You know that d/dx(e^x) is always e^x so it'll always be there

So, first by differentiating top and bottom I get..

[(e^x)-1]/(3xÂ²)

Second time differentiating I get..

(e^x)/6x

Third time differentiating I get

(e^x)/6  And here you can plug 0 for x and you get

(e^0)/6 which is..

1/6

Matthew · Answer

L'Hospital's rule can only be applied when the limit exists, so it cannot be applied in this case, as the function is divergent (i.e - the limit does not exist).

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Limits with indeterminant form, using L'Hopital's rule, help?

3 Answers