Mixing
$lim_xto 0 x^alphae^=0$
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Given any $alpha > 0$, I need to show that for $ x in [0,infty)$
beginequation
lim_xto 0 x^alphae^^1/2=0
endequation
I have tried using L'Hospital's rule. But I am not able to arrive at answer.
Thank you in advance.
real-analysis limits
asked 2 days ago
Rahul Raju Pattar
345110
add a comment |Â
up vote
0
down vote
favorite
Given any $alpha > 0$, I need to show that for $ x in [0,infty)$
beginequation
lim_xto 0 x^alphae^^1/2=0
endequation
I have tried using L'Hospital's rule. But I am not able to arrive at answer.
Thank you in advance.
real-analysis limits
asked 2 days ago
Rahul Raju Pattar
345110
Please explain in detail how it is you "tried using L'Hospital's rule." On its face this expression is not a ratio, so L'Hospital's rule will not be immediately applicable. Also, it is not a good practice to create a title that consists entirely of $LaTeX$.
– hardmath
2 days ago
@hardmath: we can write it as a ratio with exponential term in the denominator.
– Rahul Raju Pattar
2 days ago
What is $log x$ for $x<0?$
– zhw.
2 days ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given any $alpha > 0$, I need to show that for $ x in [0,infty)$
beginequation
lim_xto 0 x^alphae^^1/2=0
endequation
I have tried using L'Hospital's rule. But I am not able to arrive at answer.
Thank you in advance.
real-analysis limits
asked 2 days ago
Rahul Raju Pattar
345110
Given any $alpha > 0$, I need to show that for $ x in [0,infty)$
beginequation
lim_xto 0 x^alphae^^1/2=0
endequation
I have tried using L'Hospital's rule. But I am not able to arrive at answer.
Thank you in advance.
real-analysis limits
asked 2 days ago
Rahul Raju Pattar
345110
edited 2 days ago
asked 2 days ago
Rahul Raju Pattar
345110
asked 2 days ago
Rahul Raju Pattar
345110
asked 2 days ago
Rahul Raju Pattar
345110
345110
Please explain in detail how it is you "tried using L'Hospital's rule." On its face this expression is not a ratio, so L'Hospital's rule will not be immediately applicable. Also, it is not a good practice to create a title that consists entirely of $LaTeX$.
– hardmath
2 days ago
@hardmath: we can write it as a ratio with exponential term in the denominator.
– Rahul Raju Pattar
2 days ago
What is $log x$ for $x<0?$
– zhw.
2 days ago
add a comment |Â
Please explain in detail how it is you "tried using L'Hospital's rule." On its face this expression is not a ratio, so L'Hospital's rule will not be immediately applicable. Also, it is not a good practice to create a title that consists entirely of $LaTeX$.
– hardmath
2 days ago
@hardmath: we can write it as a ratio with exponential term in the denominator.
– Rahul Raju Pattar
2 days ago
What is $log x$ for $x<0?$
– zhw.
2 days ago
Please explain in detail how it is you "tried using L'Hospital's rule." On its face this expression is not a ratio, so L'Hospital's rule will not be immediately applicable. Also, it is not a good practice to create a title that consists entirely of $LaTeX$.
– hardmath
2 days ago
Please explain in detail how it is you "tried using L'Hospital's rule." On its face this expression is not a ratio, so L'Hospital's rule will not be immediately applicable. Also, it is not a good practice to create a title that consists entirely of $LaTeX$.
– hardmath
2 days ago
@hardmath: we can write it as a ratio with exponential term in the denominator.
– Rahul Raju Pattar
2 days ago
@hardmath: we can write it as a ratio with exponential term in the denominator.
– Rahul Raju Pattar
2 days ago
What is $log x$ for $x<0?$
– zhw.
2 days ago
What is $log x$ for $x<0?$
– zhw.
2 days ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Let
$f(x)
=x^ae^^1/2
$.
$ln f(x)
=aln x+|log x|^1/2
$.
Let
$x = 1/y$,
so
$y to infty$
as
$x to 0$.
$ln f(1/y)
=aln (1/y)+|log (1/y)|^1/2
=-aln (y)+|log y|^1/2
$.
The key is that
$dfrac^1/2ln(y)
to 0$
as $y to infty$.
Therefore
$ln f(1/y)
=ln (y)(-a +dfrac^1/2ln(y))
$.
Since
$dfrac^1/2ln(y)
to 0$
as $y to infty$
and $a > 0$,
$-a +dfrac^1/2ln(y)
lt -a/2$
for large enough $y$
so that
$ln f(1/y)
to -infty$
so
$f(1/y) to 0$.
Note that this works
for any exponent
less than $1$,
not just
$frac12$.
answered 2 days ago
marty cohen
69k446122
"Let $x = 1/y$, so $y to infty$ as $x to 0$." Not so, we need $xto 0^+.$
– zhw.
2 days ago
Then $y to infty^+$.
– marty cohen
2 days ago
add a comment |Â
up vote
0
down vote
Let $$f(x) = x^alpha e^sqrtvert log x vert$$
Consider
$$log f(x) = alpha log x + sqrtvert log x vert = log x (alpha + frac sqrtvert log x vertlog x )=log x (alpha - frac sqrtvert log x vertsqrtvert log x vertsqrtvert log x vert )$$
which is
$$log f(x) = log x (alpha - frac 1sqrtvert log x vert )$$
As $x$ goes to zero $sqrtvert log x vert$ goes to $+ infty$ hence $alpha - frac 1sqrtvert log x vert$ goes to $alpha$. Hence
$$lim_x rightarrow 0log f(x) =lim_x rightarrow 0 alpha log x = -infty$$
So
$$lim_x rightarrow 0 f(x) = e^-infty = 0$$
answered 2 days ago
Ahmad Bazzi
2,162417
add a comment |Â
up vote
0
down vote
Let $x=e^-yto 0$ as $yto infty$ then
$$largex^alphae^^1/2=e^-alpha y,e^sqrt y=e^sqrt y,-,alpha yto 0$$
indeed
$$sqrt y-alpha y=yleft(frac sqrt yy -alpharight)=yleft(frac 1 sqrt y-alpharight) to -infty$$
answered 2 days ago
gimusi
63.6k73480
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let
$f(x)
=x^ae^^1/2
$.
$ln f(x)
=aln x+|log x|^1/2
$.
Let
$x = 1/y$,
so
$y to infty$
as
$x to 0$.
$ln f(1/y)
=aln (1/y)+|log (1/y)|^1/2
=-aln (y)+|log y|^1/2
$.
The key is that
$dfrac^1/2ln(y)
to 0$
as $y to infty$.
Therefore
$ln f(1/y)
=ln (y)(-a +dfrac^1/2ln(y))
$.
Since
$dfrac^1/2ln(y)
to 0$
as $y to infty$
and $a > 0$,
$-a +dfrac^1/2ln(y)
lt -a/2$
for large enough $y$
so that
$ln f(1/y)
to -infty$
so
$f(1/y) to 0$.
Note that this works
for any exponent
less than $1$,
not just
$frac12$.
answered 2 days ago
marty cohen
69k446122
"Let $x = 1/y$, so $y to infty$ as $x to 0$." Not so, we need $xto 0^+.$
– zhw.
2 days ago
Then $y to infty^+$.
– marty cohen
2 days ago
add a comment |Â
up vote
1
down vote
accepted
Let
$f(x)
=x^ae^^1/2
$.
$ln f(x)
=aln x+|log x|^1/2
$.
Let
$x = 1/y$,
so
$y to infty$
as
$x to 0$.
$ln f(1/y)
=aln (1/y)+|log (1/y)|^1/2
=-aln (y)+|log y|^1/2
$.
The key is that
$dfrac^1/2ln(y)
to 0$
as $y to infty$.
Therefore
$ln f(1/y)
=ln (y)(-a +dfrac^1/2ln(y))
$.
Since
$dfrac^1/2ln(y)
to 0$
as $y to infty$
and $a > 0$,
$-a +dfrac^1/2ln(y)
lt -a/2$
for large enough $y$
so that
$ln f(1/y)
to -infty$
so
$f(1/y) to 0$.
Note that this works
for any exponent
less than $1$,
not just
$frac12$.
answered 2 days ago
marty cohen
69k446122
"Let $x = 1/y$, so $y to infty$ as $x to 0$." Not so, we need $xto 0^+.$
– zhw.
2 days ago
Then $y to infty^+$.
– marty cohen
2 days ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let
$f(x)
=x^ae^^1/2
$.
$ln f(x)
=aln x+|log x|^1/2
$.
Let
$x = 1/y$,
so
$y to infty$
as
$x to 0$.
$ln f(1/y)
=aln (1/y)+|log (1/y)|^1/2
=-aln (y)+|log y|^1/2
$.
The key is that
$dfrac^1/2ln(y)
to 0$
as $y to infty$.
Therefore
$ln f(1/y)
=ln (y)(-a +dfrac^1/2ln(y))
$.
Since
$dfrac^1/2ln(y)
to 0$
as $y to infty$
and $a > 0$,
$-a +dfrac^1/2ln(y)
lt -a/2$
for large enough $y$
so that
$ln f(1/y)
to -infty$
so
$f(1/y) to 0$.
Note that this works
for any exponent
less than $1$,
not just
$frac12$.
answered 2 days ago
marty cohen
69k446122
Let
$f(x)
=x^ae^^1/2
$.
$ln f(x)
=aln x+|log x|^1/2
$.
Let
$x = 1/y$,
so
$y to infty$
as
$x to 0$.
$ln f(1/y)
=aln (1/y)+|log (1/y)|^1/2
=-aln (y)+|log y|^1/2
$.
The key is that
$dfrac^1/2ln(y)
to 0$
as $y to infty$.
Therefore
$ln f(1/y)
=ln (y)(-a +dfrac^1/2ln(y))
$.
Since
$dfrac^1/2ln(y)
to 0$
as $y to infty$
and $a > 0$,
$-a +dfrac^1/2ln(y)
lt -a/2$
for large enough $y$
so that
$ln f(1/y)
to -infty$
so
$f(1/y) to 0$.
Note that this works
for any exponent
less than $1$,
not just
$frac12$.
answered 2 days ago
marty cohen
69k446122
answered 2 days ago
marty cohen
69k446122
answered 2 days ago
marty cohen
69k446122
answered 2 days ago
marty cohen
69k446122
69k446122
"Let $x = 1/y$, so $y to infty$ as $x to 0$." Not so, we need $xto 0^+.$
– zhw.
2 days ago
Then $y to infty^+$.
– marty cohen
2 days ago
add a comment |Â
"Let $x = 1/y$, so $y to infty$ as $x to 0$." Not so, we need $xto 0^+.$
– zhw.
2 days ago
Then $y to infty^+$.
– marty cohen
2 days ago
"Let $x = 1/y$, so $y to infty$ as $x to 0$." Not so, we need $xto 0^+.$
– zhw.
2 days ago
"Let $x = 1/y$, so $y to infty$ as $x to 0$." Not so, we need $xto 0^+.$
– zhw.
2 days ago
Then $y to infty^+$.
– marty cohen
2 days ago
Then $y to infty^+$.
– marty cohen
2 days ago
add a comment |Â
up vote
0
down vote
Let $$f(x) = x^alpha e^sqrtvert log x vert$$
Consider
$$log f(x) = alpha log x + sqrtvert log x vert = log x (alpha + frac sqrtvert log x vertlog x )=log x (alpha - frac sqrtvert log x vertsqrtvert log x vertsqrtvert log x vert )$$
which is
$$log f(x) = log x (alpha - frac 1sqrtvert log x vert )$$
As $x$ goes to zero $sqrtvert log x vert$ goes to $+ infty$ hence $alpha - frac 1sqrtvert log x vert$ goes to $alpha$. Hence
$$lim_x rightarrow 0log f(x) =lim_x rightarrow 0 alpha log x = -infty$$
So
$$lim_x rightarrow 0 f(x) = e^-infty = 0$$
answered 2 days ago
Ahmad Bazzi
2,162417
add a comment |Â
up vote
0
down vote
Let $$f(x) = x^alpha e^sqrtvert log x vert$$
Consider
$$log f(x) = alpha log x + sqrtvert log x vert = log x (alpha + frac sqrtvert log x vertlog x )=log x (alpha - frac sqrtvert log x vertsqrtvert log x vertsqrtvert log x vert )$$
which is
$$log f(x) = log x (alpha - frac 1sqrtvert log x vert )$$
As $x$ goes to zero $sqrtvert log x vert$ goes to $+ infty$ hence $alpha - frac 1sqrtvert log x vert$ goes to $alpha$. Hence
$$lim_x rightarrow 0log f(x) =lim_x rightarrow 0 alpha log x = -infty$$
So
$$lim_x rightarrow 0 f(x) = e^-infty = 0$$
answered 2 days ago
Ahmad Bazzi
2,162417
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $$f(x) = x^alpha e^sqrtvert log x vert$$
Consider
$$log f(x) = alpha log x + sqrtvert log x vert = log x (alpha + frac sqrtvert log x vertlog x )=log x (alpha - frac sqrtvert log x vertsqrtvert log x vertsqrtvert log x vert )$$
which is
$$log f(x) = log x (alpha - frac 1sqrtvert log x vert )$$
As $x$ goes to zero $sqrtvert log x vert$ goes to $+ infty$ hence $alpha - frac 1sqrtvert log x vert$ goes to $alpha$. Hence
$$lim_x rightarrow 0log f(x) =lim_x rightarrow 0 alpha log x = -infty$$
So
$$lim_x rightarrow 0 f(x) = e^-infty = 0$$
answered 2 days ago
Ahmad Bazzi
2,162417
Let $$f(x) = x^alpha e^sqrtvert log x vert$$
Consider
$$log f(x) = alpha log x + sqrtvert log x vert = log x (alpha + frac sqrtvert log x vertlog x )=log x (alpha - frac sqrtvert log x vertsqrtvert log x vertsqrtvert log x vert )$$
which is
$$log f(x) = log x (alpha - frac 1sqrtvert log x vert )$$
As $x$ goes to zero $sqrtvert log x vert$ goes to $+ infty$ hence $alpha - frac 1sqrtvert log x vert$ goes to $alpha$. Hence
$$lim_x rightarrow 0log f(x) =lim_x rightarrow 0 alpha log x = -infty$$
So
$$lim_x rightarrow 0 f(x) = e^-infty = 0$$
answered 2 days ago
Ahmad Bazzi
2,162417
edited 2 days ago
answered 2 days ago
Ahmad Bazzi
2,162417
answered 2 days ago
Ahmad Bazzi
2,162417
answered 2 days ago
Ahmad Bazzi
2,162417
2,162417
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $x=e^-yto 0$ as $yto infty$ then
$$largex^alphae^^1/2=e^-alpha y,e^sqrt y=e^sqrt y,-,alpha yto 0$$
indeed
$$sqrt y-alpha y=yleft(frac sqrt yy -alpharight)=yleft(frac 1 sqrt y-alpharight) to -infty$$
answered 2 days ago
gimusi
63.6k73480
add a comment |Â
up vote
0
down vote
Let $x=e^-yto 0$ as $yto infty$ then
$$largex^alphae^^1/2=e^-alpha y,e^sqrt y=e^sqrt y,-,alpha yto 0$$
indeed
$$sqrt y-alpha y=yleft(frac sqrt yy -alpharight)=yleft(frac 1 sqrt y-alpharight) to -infty$$
answered 2 days ago
gimusi
63.6k73480
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $x=e^-yto 0$ as $yto infty$ then
$$largex^alphae^^1/2=e^-alpha y,e^sqrt y=e^sqrt y,-,alpha yto 0$$
indeed
$$sqrt y-alpha y=yleft(frac sqrt yy -alpharight)=yleft(frac 1 sqrt y-alpharight) to -infty$$
answered 2 days ago
gimusi
63.6k73480
Let $x=e^-yto 0$ as $yto infty$ then
$$largex^alphae^^1/2=e^-alpha y,e^sqrt y=e^sqrt y,-,alpha yto 0$$
indeed
$$sqrt y-alpha y=yleft(frac sqrt yy -alpharight)=yleft(frac 1 sqrt y-alpharight) to -infty$$
answered 2 days ago
gimusi
63.6k73480
answered 2 days ago
gimusi
63.6k73480
answered 2 days ago
gimusi
63.6k73480
answered 2 days ago
gimusi
63.6k73480
63.6k73480
add a comment |Â
add a comment |Â
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Please explain in detail how it is you "tried using L'Hospital's rule." On its face this expression is not a ratio, so L'Hospital's rule will not be immediately applicable. Also, it is not a good practice to create a title that consists entirely of $LaTeX$.
– hardmath
2 days ago
@hardmath: we can write it as a ratio with exponential term in the denominator.
– Rahul Raju Pattar
2 days ago
What is $log x$ for $x<0?$
– zhw.
2 days ago