Mixing
Calculating limit of ln(arctan(x)) using chain rule
$begingroup$
I know there are other options to do that, but I'm trying to use the chain rule to calculate this limit and I get a wrong answer.
So what's wrong in the following calculation:
$limlimits_{x to 0+} ln(arctan(x)) = frac{1}{(1+x^2)} * frac{1}{arctan(x)} = frac{1}{0+} = infty $
calculus derivatives
edited Jan 13 at 22:40
user
4,4201929
asked Jan 13 at 22:12
RizonRizon
1095
$endgroup$
add a comment |
$begingroup$
I know there are other options to do that, but I'm trying to use the chain rule to calculate this limit and I get a wrong answer.
So what's wrong in the following calculation:
$limlimits_{x to 0+} ln(arctan(x)) = frac{1}{(1+x^2)} * frac{1}{arctan(x)} = frac{1}{0+} = infty $
calculus derivatives
edited Jan 13 at 22:40
user
4,4201929
asked Jan 13 at 22:12
RizonRizon
1095
$endgroup$
2
$begingroup$
You can't just take the derivative and expect to get the same limit.
$endgroup$
– Matt Samuel
Jan 13 at 22:18
add a comment |
$begingroup$
I know there are other options to do that, but I'm trying to use the chain rule to calculate this limit and I get a wrong answer.
So what's wrong in the following calculation:
$limlimits_{x to 0+} ln(arctan(x)) = frac{1}{(1+x^2)} * frac{1}{arctan(x)} = frac{1}{0+} = infty $
calculus derivatives
edited Jan 13 at 22:40
user
4,4201929
asked Jan 13 at 22:12
RizonRizon
1095
$endgroup$
I know there are other options to do that, but I'm trying to use the chain rule to calculate this limit and I get a wrong answer.
So what's wrong in the following calculation:
$limlimits_{x to 0+} ln(arctan(x)) = frac{1}{(1+x^2)} * frac{1}{arctan(x)} = frac{1}{0+} = infty $
calculus derivatives
calculus derivatives
edited Jan 13 at 22:40
user
4,4201929
asked Jan 13 at 22:12
RizonRizon
1095
edited Jan 13 at 22:40
user
4,4201929
asked Jan 13 at 22:12
RizonRizon
1095
edited Jan 13 at 22:40
user
4,4201929
edited Jan 13 at 22:40
user
4,4201929
edited Jan 13 at 22:40
user
4,4201929
4,4201929
asked Jan 13 at 22:12
RizonRizon
1095
asked Jan 13 at 22:12
RizonRizon
1095
asked Jan 13 at 22:12
RizonRizon
1095
1095
2
$begingroup$
You can't just take the derivative and expect to get the same limit.
$endgroup$
– Matt Samuel
Jan 13 at 22:18
add a comment |
2
$begingroup$
You can't just take the derivative and expect to get the same limit.
$endgroup$
– Matt Samuel
Jan 13 at 22:18
2
2
$begingroup$
You can't just take the derivative and expect to get the same limit.
$endgroup$
– Matt Samuel
Jan 13 at 22:18
$begingroup$
You can't just take the derivative and expect to get the same limit.
$endgroup$
– Matt Samuel
Jan 13 at 22:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try this with the function $2x$.
$$lim_{xto 0} 2x =0.$$
Now take the derivative of $2x$, which is $2$:
$$lim_{xto 0} 2 = 2.$$
Hmmmm....different answers. This may be a L'hospital's Rule confusion.
answered Jan 13 at 22:25
B. GoddardB. Goddard
18.9k21440
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Try this with the function $2x$.
$$lim_{xto 0} 2x =0.$$
Now take the derivative of $2x$, which is $2$:
$$lim_{xto 0} 2 = 2.$$
Hmmmm....different answers. This may be a L'hospital's Rule confusion.
answered Jan 13 at 22:25
B. GoddardB. Goddard
18.9k21440
$endgroup$
add a comment |
$begingroup$
Try this with the function $2x$.
$$lim_{xto 0} 2x =0.$$
Now take the derivative of $2x$, which is $2$:
$$lim_{xto 0} 2 = 2.$$
Hmmmm....different answers. This may be a L'hospital's Rule confusion.
answered Jan 13 at 22:25
B. GoddardB. Goddard
18.9k21440
$endgroup$
add a comment |
$begingroup$
Try this with the function $2x$.
$$lim_{xto 0} 2x =0.$$
Now take the derivative of $2x$, which is $2$:
$$lim_{xto 0} 2 = 2.$$
Hmmmm....different answers. This may be a L'hospital's Rule confusion.
answered Jan 13 at 22:25
B. GoddardB. Goddard
18.9k21440
$endgroup$
Try this with the function $2x$.
$$lim_{xto 0} 2x =0.$$
Now take the derivative of $2x$, which is $2$:
$$lim_{xto 0} 2 = 2.$$
Hmmmm....different answers. This may be a L'hospital's Rule confusion.
answered Jan 13 at 22:25
B. GoddardB. Goddard
18.9k21440
answered Jan 13 at 22:25
B. GoddardB. Goddard
18.9k21440
answered Jan 13 at 22:25
B. GoddardB. Goddard
18.9k21440
answered Jan 13 at 22:25
B. GoddardB. Goddard
18.9k21440
18.9k21440
add a comment |
add a comment |
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$begingroup$
You can't just take the derivative and expect to get the same limit.
$endgroup$
– Matt Samuel
Jan 13 at 22:18