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L'Hopital Rule (1 Viewer)
- Thread starter AlienC
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I never have done this but I can comment on this and this is that in each stage they find the first derivative of bottom and top. For example first at top you have e^x-1-x then in the next level you have derivative of this which is e^x-1 and in the next level you have derivative of this which is e^x. you can follow the same rule for bottom first you have x^2 then you have derivative of it which is 2x in the next level. in the last level you again have derivative of that which is 2.
Slidey
But pieces of what?
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Alwpays remember that you canp onlyp aplyp Lp'Hopsptial's prulpe when both thpe bottom and thpe top eachp go towards pzero whpen you take their lpimits.p If one of them does not, then you cannot aplyp the rulep.
Pardon the p's. This keypboard is brokenp.
Pardon the p's. This keypboard is brokenp.
It's just a way to find limits when simple evaluation results in 0/0, inf/inf or -inf/inf.
Eg. lim (x-> inf) e^x/x is inf/inf (which is not equal to 1). The e^x overpowers the x, which can be seen by applying l'hopital's rule:
lim (x-> inf) e^x/x = lim (x-> inf) e^x/1 = lim (x-> inf) e^x = inf
And yeah, its generally met first year of uni, but its not difficult to learn earlier (its proof requires uni maths though)
Eg. lim (x-> inf) e^x/x is inf/inf (which is not equal to 1). The e^x overpowers the x, which can be seen by applying l'hopital's rule:
lim (x-> inf) e^x/x = lim (x-> inf) e^x/1 = lim (x-> inf) e^x = inf
And yeah, its generally met first year of uni, but its not difficult to learn earlier (its proof requires uni maths though)
Trev
stix
Hahaha, Slide...Slidey said:
darkliight
I ponder, weak and weary
Not the bolded one, iirc.acmilan said:
I figure if lim (x->a) f(x)/g(x) = -inf/inf, then you can just convert it to -lim (x->a) -f(x)/g(x), in which case the limit becomes inf/inf, with a negative on the outside.darkliight said:
Eg: lim (x->inf) -ex/x is of the form -inf/inf (l'hopitals rule works in this case without removing the - sign), but just remove it anyways to get -lim (x->inf) ex/x, the limit is now in the inf/inf form and you can apply the rule to get -inf as the answer.