the question says find this limit using delhospitals rule
i got stack here, i think i did something wrong with the derevatives, but the method bringing e to the denominator was the method the teacher told us to use from there i had some problems..
i got stack here, i think i did something wrong with the derevatives, but the method bringing e to the denominator was the method the teacher told us to use from there i had some problems..
(edited 7 months ago)
For the denominator, after lhopital, you could write y=1/x, then the denom becomes
y^2 e^(-y)
as x->0, y->inf and it should be fairly clear that the exponential will dominate, so the denom -> 0 and the fraction tends to
1/0
so a vertical asymptote for the original function (as youd probably expect).
y^2 e^(-y)
as x->0, y->inf and it should be fairly clear that the exponential will dominate, so the denom -> 0 and the fraction tends to
1/0
so a vertical asymptote for the original function (as youd probably expect).
Original post by mqb2766
For the denominator, after lhopital, you could write y=1/x, then the denom becomes
y^2 e^(-y)
as x->0, y->inf and it should be fairly clear that the exponential will dominate, so the denom -> 0 and the fraction tends to
1/0
so a vertical asymptote for the original function (as youd probably expect).
y^2 e^(-y)
as x->0, y->inf and it should be fairly clear that the exponential will dominate, so the denom -> 0 and the fraction tends to
1/0
so a vertical asymptote for the original function (as youd probably expect).
but i am asked the answer, the value to the limit
so the answer of the limit is +infinity?
Original post by mikeft
but i am asked the answer, the value to the limit
so the answer of the limit is +infinity?
so the answer of the limit is +infinity?
The function does tend to +inf as x->0+
https://www.desmos.com/calculator/hyfcxnou9g
Second thoughts, not sure my reasoning would be that good as its similar to the original function, though it has the germ of the idea. Instead try doing lhopital on
e^y / y
where y=1/x and y->inf and that should give.
(edited 7 months ago)
Original post by mqb2766
The function does tend to +inf as x->0+
https://www.desmos.com/calculator/hyfcxnou9g
Second thoughts, not sure my reasoning would be that good as its similar to the original function, though it has the germ of the idea. Instead try doing lhopital on
e^y / y
where y=1/x and y->inf and that should give.
https://www.desmos.com/calculator/hyfcxnou9g
Second thoughts, not sure my reasoning would be that good as its similar to the original function, though it has the germ of the idea. Instead try doing lhopital on
e^y / y
where y=1/x and y->inf and that should give.
i am not sure i have understood the e to the y stuff you’ve written..
the answer to the limit is infinity? and is my work correct? because i think i got lost at some point.
Original post by mikeft
i am not sure i have understood the e to the y stuff you’ve written..
the answer to the limit is infinity? and is my work correct? because i think i got lost at some point.
the answer to the limit is infinity? and is my work correct? because i think i got lost at some point.
Just transform the problem using y=1/x so
xe^(1/x) = e^y/y
and as x->0+, then y->inf. So do lhopital on
e^y/y
as y->inf.
The 0/0 is wrong for your lhopital, it should be 1/0, but you have to argue that exponentals decay faster than a quadratic (they do, but the above is simpler).
(edited 7 months ago)
To be somewhat "meta" about this problem.
The reason why the limit x exp(1/x) isn't obvious is that x is going to 0 but exp(1/x) is going to infinity.
It's difficult to solve this using L'Hopital directly because after you transform to x/exp(-1/x), differentiating leaves you considering x^2/exp(-1/x), which is harder to deal with than you started with.
You can't reasonably argue "exponentials beat powers" (although it's true) because the original problem is "prove a particular case of exponentials beat powers". To me it's "begging the question" even if it's not exactly circular.
As mqb says, substituting y=1/x, so that differentiating the exponential doesn't give you something that grows exponentially more complicated, is the way to go here.
Generally speaking, if a method requires you to differentiate f(x^n) (with n not 1) more than once, think carefully about whether there's a better option.
The reason why the limit x exp(1/x) isn't obvious is that x is going to 0 but exp(1/x) is going to infinity.
It's difficult to solve this using L'Hopital directly because after you transform to x/exp(-1/x), differentiating leaves you considering x^2/exp(-1/x), which is harder to deal with than you started with.
You can't reasonably argue "exponentials beat powers" (although it's true) because the original problem is "prove a particular case of exponentials beat powers". To me it's "begging the question" even if it's not exactly circular.
As mqb says, substituting y=1/x, so that differentiating the exponential doesn't give you something that grows exponentially more complicated, is the way to go here.
Generally speaking, if a method requires you to differentiate f(x^n) (with n not 1) more than once, think carefully about whether there's a better option.
(edited 7 months ago)
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