Mixing
Indetermiante form limits question.
$begingroup$
I have a simple question. I learned these things years ago but cannot recall how to prove this. I am interested in the right hand limit
$$lim_{x rightarrow 0^{+}} frac{1}{sqrt{x}} e^{-tfrac{a}{2x}}$$
where $a$ is just some constant different to zero.
My intuition is that this goes to zero as the exponential decays much faster than the square root.
I tried to use L'Hôpital but it didn't help. Did I make a mistake or do I just need to keep applying it over and over again.
begin{gather}
f(x) = e^{-frac{a}{2x}}
\
g(x) = sqrt{x}
\
frac{f'(x)}{g'(x)} = -(frac{a}{2x^2}e^{-frac{a}{2x}})(2sqrt{x})
end{gather}
Am I just missing something? I feel like I am forgetting something very very basic. I also considered expanding the exponential in Taylor series but that didn't really help
limits
edited Jan 11 at 16:19
egreg
181k1485203
asked Jan 11 at 15:54
TCDTCD
33
$endgroup$
add a comment |
$begingroup$
I have a simple question. I learned these things years ago but cannot recall how to prove this. I am interested in the right hand limit
$$lim_{x rightarrow 0^{+}} frac{1}{sqrt{x}} e^{-tfrac{a}{2x}}$$
where $a$ is just some constant different to zero.
My intuition is that this goes to zero as the exponential decays much faster than the square root.
I tried to use L'Hôpital but it didn't help. Did I make a mistake or do I just need to keep applying it over and over again.
begin{gather}
f(x) = e^{-frac{a}{2x}}
\
g(x) = sqrt{x}
\
frac{f'(x)}{g'(x)} = -(frac{a}{2x^2}e^{-frac{a}{2x}})(2sqrt{x})
end{gather}
Am I just missing something? I feel like I am forgetting something very very basic. I also considered expanding the exponential in Taylor series but that didn't really help
limits
edited Jan 11 at 16:19
egreg
181k1485203
asked Jan 11 at 15:54
TCDTCD
33
$endgroup$
1
$begingroup$
Make the substitution $x=1/y$.
$endgroup$
– Mindlack
Jan 11 at 15:57
$begingroup$
I suppose $a$ is a positive non-zero constant?
$endgroup$
– StackTD
Jan 11 at 16:03
$begingroup$
Thanks @Mindlack. I think that works out ok. Is that a standard kind of technique or did you just know from practice? .
$endgroup$
– TCD
Jan 11 at 16:07
$begingroup$
@StackTD. Yes, we can assume a positive
$endgroup$
– TCD
Jan 11 at 16:08
$begingroup$
TCD: I wouldn’t say it is a standard technique because in general, substitutions in limits are little more than rewriting stuff with no change in viewpoint (the expression can definitely look simpler, but it is sort of artificial, we just split the complexity). Moreover, that’s a rather standard example, so that is practice (try and understand what happens to all the derivatives of the function when the argument goes to $0$).
$endgroup$
– Mindlack
Jan 11 at 16:19
add a comment |
$begingroup$
I have a simple question. I learned these things years ago but cannot recall how to prove this. I am interested in the right hand limit
$$lim_{x rightarrow 0^{+}} frac{1}{sqrt{x}} e^{-tfrac{a}{2x}}$$
where $a$ is just some constant different to zero.
My intuition is that this goes to zero as the exponential decays much faster than the square root.
I tried to use L'Hôpital but it didn't help. Did I make a mistake or do I just need to keep applying it over and over again.
begin{gather}
f(x) = e^{-frac{a}{2x}}
\
g(x) = sqrt{x}
\
frac{f'(x)}{g'(x)} = -(frac{a}{2x^2}e^{-frac{a}{2x}})(2sqrt{x})
end{gather}
Am I just missing something? I feel like I am forgetting something very very basic. I also considered expanding the exponential in Taylor series but that didn't really help
limits
edited Jan 11 at 16:19
egreg
181k1485203
asked Jan 11 at 15:54
TCDTCD
33
$endgroup$
I have a simple question. I learned these things years ago but cannot recall how to prove this. I am interested in the right hand limit
$$lim_{x rightarrow 0^{+}} frac{1}{sqrt{x}} e^{-tfrac{a}{2x}}$$
where $a$ is just some constant different to zero.
My intuition is that this goes to zero as the exponential decays much faster than the square root.
I tried to use L'Hôpital but it didn't help. Did I make a mistake or do I just need to keep applying it over and over again.
begin{gather}
f(x) = e^{-frac{a}{2x}}
\
g(x) = sqrt{x}
\
frac{f'(x)}{g'(x)} = -(frac{a}{2x^2}e^{-frac{a}{2x}})(2sqrt{x})
end{gather}
Am I just missing something? I feel like I am forgetting something very very basic. I also considered expanding the exponential in Taylor series but that didn't really help
limits
limits
edited Jan 11 at 16:19
egreg
181k1485203
asked Jan 11 at 15:54
TCDTCD
33
edited Jan 11 at 16:19
egreg
181k1485203
asked Jan 11 at 15:54
TCDTCD
33
edited Jan 11 at 16:19
egreg
181k1485203
edited Jan 11 at 16:19
egreg
181k1485203
edited Jan 11 at 16:19
egreg
181k1485203
181k1485203
asked Jan 11 at 15:54
TCDTCD
33
asked Jan 11 at 15:54
TCDTCD
33
asked Jan 11 at 15:54
TCDTCD
33
33
1
$begingroup$
Make the substitution $x=1/y$.
$endgroup$
– Mindlack
Jan 11 at 15:57
$begingroup$
I suppose $a$ is a positive non-zero constant?
$endgroup$
– StackTD
Jan 11 at 16:03
$begingroup$
Thanks @Mindlack. I think that works out ok. Is that a standard kind of technique or did you just know from practice? .
$endgroup$
– TCD
Jan 11 at 16:07
$begingroup$
@StackTD. Yes, we can assume a positive
$endgroup$
– TCD
Jan 11 at 16:08
$begingroup$
TCD: I wouldn’t say it is a standard technique because in general, substitutions in limits are little more than rewriting stuff with no change in viewpoint (the expression can definitely look simpler, but it is sort of artificial, we just split the complexity). Moreover, that’s a rather standard example, so that is practice (try and understand what happens to all the derivatives of the function when the argument goes to $0$).
$endgroup$
– Mindlack
Jan 11 at 16:19
add a comment |
1
$begingroup$
Make the substitution $x=1/y$.
$endgroup$
– Mindlack
Jan 11 at 15:57
$begingroup$
I suppose $a$ is a positive non-zero constant?
$endgroup$
– StackTD
Jan 11 at 16:03
$begingroup$
Thanks @Mindlack. I think that works out ok. Is that a standard kind of technique or did you just know from practice? .
$endgroup$
– TCD
Jan 11 at 16:07
$begingroup$
@StackTD. Yes, we can assume a positive
$endgroup$
– TCD
Jan 11 at 16:08
$begingroup$
TCD: I wouldn’t say it is a standard technique because in general, substitutions in limits are little more than rewriting stuff with no change in viewpoint (the expression can definitely look simpler, but it is sort of artificial, we just split the complexity). Moreover, that’s a rather standard example, so that is practice (try and understand what happens to all the derivatives of the function when the argument goes to $0$).
$endgroup$
– Mindlack
Jan 11 at 16:19
1
1
$begingroup$
Make the substitution $x=1/y$.
$endgroup$
– Mindlack
Jan 11 at 15:57
$begingroup$
Make the substitution $x=1/y$.
$endgroup$
– Mindlack
Jan 11 at 15:57
$begingroup$
I suppose $a$ is a positive non-zero constant?
$endgroup$
– StackTD
Jan 11 at 16:03
$begingroup$
I suppose $a$ is a positive non-zero constant?
$endgroup$
– StackTD
Jan 11 at 16:03
$begingroup$
Thanks @Mindlack. I think that works out ok. Is that a standard kind of technique or did you just know from practice? .
$endgroup$
– TCD
Jan 11 at 16:07
$begingroup$
Thanks @Mindlack. I think that works out ok. Is that a standard kind of technique or did you just know from practice? .
$endgroup$
– TCD
Jan 11 at 16:07
$begingroup$
@StackTD. Yes, we can assume a positive
$endgroup$
– TCD
Jan 11 at 16:08
$begingroup$
@StackTD. Yes, we can assume a positive
$endgroup$
– TCD
Jan 11 at 16:08
$begingroup$
TCD: I wouldn’t say it is a standard technique because in general, substitutions in limits are little more than rewriting stuff with no change in viewpoint (the expression can definitely look simpler, but it is sort of artificial, we just split the complexity). Moreover, that’s a rather standard example, so that is practice (try and understand what happens to all the derivatives of the function when the argument goes to $0$).
$endgroup$
– Mindlack
Jan 11 at 16:19
$begingroup$
TCD: I wouldn’t say it is a standard technique because in general, substitutions in limits are little more than rewriting stuff with no change in viewpoint (the expression can definitely look simpler, but it is sort of artificial, we just split the complexity). Moreover, that’s a rather standard example, so that is practice (try and understand what happens to all the derivatives of the function when the argument goes to $0$).
$endgroup$
– Mindlack
Jan 11 at 16:19
add a comment |
1 Answer
1
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$begingroup$
It's simpler if you do $sqrt{x}=1/t$, so the limit becomes
$$
lim_{ttoinfty}te^{-at^2/2}=
lim_{ttoinfty}frac{t}{e^{at^2/2}}
$$
answered Jan 11 at 16:21
egregegreg
181k1485203
$endgroup$
add a comment |
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$begingroup$
It's simpler if you do $sqrt{x}=1/t$, so the limit becomes
$$
lim_{ttoinfty}te^{-at^2/2}=
lim_{ttoinfty}frac{t}{e^{at^2/2}}
$$
answered Jan 11 at 16:21
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
It's simpler if you do $sqrt{x}=1/t$, so the limit becomes
$$
lim_{ttoinfty}te^{-at^2/2}=
lim_{ttoinfty}frac{t}{e^{at^2/2}}
$$
answered Jan 11 at 16:21
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
It's simpler if you do $sqrt{x}=1/t$, so the limit becomes
$$
lim_{ttoinfty}te^{-at^2/2}=
lim_{ttoinfty}frac{t}{e^{at^2/2}}
$$
answered Jan 11 at 16:21
egregegreg
181k1485203
$endgroup$
It's simpler if you do $sqrt{x}=1/t$, so the limit becomes
$$
lim_{ttoinfty}te^{-at^2/2}=
lim_{ttoinfty}frac{t}{e^{at^2/2}}
$$
answered Jan 11 at 16:21
egregegreg
181k1485203
answered Jan 11 at 16:21
egregegreg
181k1485203
answered Jan 11 at 16:21
egregegreg
181k1485203
answered Jan 11 at 16:21
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
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1
$begingroup$
Make the substitution $x=1/y$.
$endgroup$
– Mindlack
Jan 11 at 15:57
$begingroup$
I suppose $a$ is a positive non-zero constant?
$endgroup$
– StackTD
Jan 11 at 16:03
$begingroup$
Thanks @Mindlack. I think that works out ok. Is that a standard kind of technique or did you just know from practice? .
$endgroup$
– TCD
Jan 11 at 16:07
$begingroup$
@StackTD. Yes, we can assume a positive
$endgroup$
– TCD
Jan 11 at 16:08
$begingroup$
TCD: I wouldn’t say it is a standard technique because in general, substitutions in limits are little more than rewriting stuff with no change in viewpoint (the expression can definitely look simpler, but it is sort of artificial, we just split the complexity). Moreover, that’s a rather standard example, so that is practice (try and understand what happens to all the derivatives of the function when the argument goes to $0$).
$endgroup$
– Mindlack
Jan 11 at 16:19