Mixing
$lim_{n to infty}frac{e^sqrt{n}}{c^n}$?
$begingroup$
How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$
$$lim_{n to infty}dfrac{e^sqrt{n}}{c^n}$$
I'm having trouble on how to approach this problem.
limits exponential-function
edited Feb 3 at 12:53
Thomas Shelby
4,7532727
asked Feb 3 at 12:09
doobledooble
153
$endgroup$
add a comment |
$begingroup$
How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$
$$lim_{n to infty}dfrac{e^sqrt{n}}{c^n}$$
I'm having trouble on how to approach this problem.
limits exponential-function
edited Feb 3 at 12:53
Thomas Shelby
4,7532727
asked Feb 3 at 12:09
doobledooble
153
$endgroup$
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
add a comment |
$begingroup$
How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$
$$lim_{n to infty}dfrac{e^sqrt{n}}{c^n}$$
I'm having trouble on how to approach this problem.
limits exponential-function
edited Feb 3 at 12:53
Thomas Shelby
4,7532727
asked Feb 3 at 12:09
doobledooble
153
$endgroup$
How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$
$$lim_{n to infty}dfrac{e^sqrt{n}}{c^n}$$
I'm having trouble on how to approach this problem.
limits exponential-function
limits exponential-function
edited Feb 3 at 12:53
Thomas Shelby
4,7532727
asked Feb 3 at 12:09
doobledooble
153
edited Feb 3 at 12:53
Thomas Shelby
4,7532727
asked Feb 3 at 12:09
doobledooble
153
edited Feb 3 at 12:53
Thomas Shelby
4,7532727
edited Feb 3 at 12:53
Thomas Shelby
4,7532727
edited Feb 3 at 12:53
Thomas Shelby
4,7532727
4,7532727
asked Feb 3 at 12:09
doobledooble
153
asked Feb 3 at 12:09
doobledooble
153
asked Feb 3 at 12:09
doobledooble
153
153
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
add a comment |
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
$endgroup$
add a comment |
$begingroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
$endgroup$
add a comment |
$begingroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
$endgroup$
Your limit can be write as $lim e^{sqrt{n}-log(c)n}$, and $lim (sqrt{n}-log(c)n)= -infty$, so $lim e^{sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).
answered Feb 3 at 12:24
BonbonBonbon
45118
answered Feb 3 at 12:24
BonbonBonbon
45118
answered Feb 3 at 12:24
BonbonBonbon
45118
answered Feb 3 at 12:24
BonbonBonbon
45118
45118
add a comment |
add a comment |
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$begingroup$
Sketch a graph of the logarithm of your ratio on the vertical axis and $sqrt n$ on the horizontal axis. You will meet an old friend.
$endgroup$
– kimchi lover
Feb 3 at 12:11
$begingroup$
The limit is equal to zero
$endgroup$
– Dr. Sonnhard Graubner
Feb 3 at 12:16