What is the meaning of a limit as $e^{(n+1) infty}$ in WolframAlpha?
1
$begingroup$
I inputted the following into WolframAlpha:
lim x to 0 of (cotx)/x^n
And I got
$$lim_{x rightarrow 0}{frac{cot{x}}{x^n}} = e^{(n+1) infty}$$
What does this mean?
limits wolfram-alpha
edited Dec 22 '18 at 0:10
rafa11111
1,2042417
asked Dec 21 '18 at 23:53
Shrey JoshiShrey Joshi
31413
$endgroup$
add a comment |
1
$begingroup$
I inputted the following into WolframAlpha:
lim x to 0 of (cotx)/x^n
And I got
$$lim_{x rightarrow 0}{frac{cot{x}}{x^n}} = e^{(n+1) infty}$$
What does this mean?
limits wolfram-alpha
edited Dec 22 '18 at 0:10
rafa11111
1,2042417
asked Dec 21 '18 at 23:53
Shrey JoshiShrey Joshi
31413
$endgroup$
2
$begingroup$
I think if you write that as an answer on an exam, it will be marked wrong.
$endgroup$
– GEdgar
Dec 22 '18 at 1:06
$begingroup$
WA often gives Wrong Answers for limits, because its programmers anyhow use ad-hoc heuristics. I have experienced them doing that: Sometimes, the wrong answer would even appear before it is substituted by the correct answer. Other times, contradictory answers would appear. At any point in time (including in the future), you can easily come up with an infinite class of limits that WA will give the Wrong Answer to. I'll leave it as a fun exercise for anyone interested.
$endgroup$
– user21820
Dec 28 '18 at 4:00
add a comment |
1
1
1
$begingroup$
I inputted the following into WolframAlpha:
lim x to 0 of (cotx)/x^n
And I got
$$lim_{x rightarrow 0}{frac{cot{x}}{x^n}} = e^{(n+1) infty}$$
What does this mean?
limits wolfram-alpha
edited Dec 22 '18 at 0:10
rafa11111
1,2042417
asked Dec 21 '18 at 23:53
Shrey JoshiShrey Joshi
31413
$endgroup$
I inputted the following into WolframAlpha:
lim x to 0 of (cotx)/x^n
And I got
$$lim_{x rightarrow 0}{frac{cot{x}}{x^n}} = e^{(n+1) infty}$$
What does this mean?
limits wolfram-alpha
limits wolfram-alpha
edited Dec 22 '18 at 0:10
rafa11111
1,2042417
asked Dec 21 '18 at 23:53
Shrey JoshiShrey Joshi
31413
edited Dec 22 '18 at 0:10
rafa11111
1,2042417
asked Dec 21 '18 at 23:53
Shrey JoshiShrey Joshi
31413
edited Dec 22 '18 at 0:10
rafa11111
1,2042417
edited Dec 22 '18 at 0:10
rafa11111
1,2042417
edited Dec 22 '18 at 0:10
rafa11111
1,2042417
1,2042417
asked Dec 21 '18 at 23:53
Shrey JoshiShrey Joshi
31413
asked Dec 21 '18 at 23:53
Shrey JoshiShrey Joshi
31413
asked Dec 21 '18 at 23:53
Shrey JoshiShrey Joshi
31413
31413
2
$begingroup$
I think if you write that as an answer on an exam, it will be marked wrong.
$endgroup$
– GEdgar
Dec 22 '18 at 1:06
$begingroup$
WA often gives Wrong Answers for limits, because its programmers anyhow use ad-hoc heuristics. I have experienced them doing that: Sometimes, the wrong answer would even appear before it is substituted by the correct answer. Other times, contradictory answers would appear. At any point in time (including in the future), you can easily come up with an infinite class of limits that WA will give the Wrong Answer to. I'll leave it as a fun exercise for anyone interested.
$endgroup$
– user21820
Dec 28 '18 at 4:00
add a comment |
2
$begingroup$
I think if you write that as an answer on an exam, it will be marked wrong.
$endgroup$
– GEdgar
Dec 22 '18 at 1:06
$begingroup$
WA often gives Wrong Answers for limits, because its programmers anyhow use ad-hoc heuristics. I have experienced them doing that: Sometimes, the wrong answer would even appear before it is substituted by the correct answer. Other times, contradictory answers would appear. At any point in time (including in the future), you can easily come up with an infinite class of limits that WA will give the Wrong Answer to. I'll leave it as a fun exercise for anyone interested.
$endgroup$
– user21820
Dec 28 '18 at 4:00
2
2
$begingroup$
I think if you write that as an answer on an exam, it will be marked wrong.
$endgroup$
– GEdgar
Dec 22 '18 at 1:06
$begingroup$
I think if you write that as an answer on an exam, it will be marked wrong.
$endgroup$
– GEdgar
Dec 22 '18 at 1:06
$begingroup$
WA often gives Wrong Answers for limits, because its programmers anyhow use ad-hoc heuristics. I have experienced them doing that: Sometimes, the wrong answer would even appear before it is substituted by the correct answer. Other times, contradictory answers would appear. At any point in time (including in the future), you can easily come up with an infinite class of limits that WA will give the Wrong Answer to. I'll leave it as a fun exercise for anyone interested.
$endgroup$
– user21820
Dec 28 '18 at 4:00
$begingroup$
WA often gives Wrong Answers for limits, because its programmers anyhow use ad-hoc heuristics. I have experienced them doing that: Sometimes, the wrong answer would even appear before it is substituted by the correct answer. Other times, contradictory answers would appear. At any point in time (including in the future), you can easily come up with an infinite class of limits that WA will give the Wrong Answer to. I'll leave it as a fun exercise for anyone interested.
$endgroup$
– user21820
Dec 28 '18 at 4:00
add a comment |
1 Answer
1
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oldest
votes
2
$begingroup$
It probably denotes that the final value of the limit depends on $n$. Specifically, if $n+1 < 0$, then the limit is $to e^{-infty} = 0$ while if $n+1 >0$, then the limit is $to e^{+infty} = infty$.
answered Dec 21 '18 at 23:58
RebellosRebellos
15.7k31250
$endgroup$
$begingroup$
But for certain values of $n$ if the limit is not defined then how does it compensate for that?
$endgroup$
– Shrey Joshi
Dec 21 '18 at 23:59
$begingroup$
And if $n+1=0$ it is $limlimits_{x rightarrow 0}{x ,{cot{x}}} = e^0=1$
$endgroup$
– Henry
Dec 22 '18 at 0:00
$begingroup$
Simple answer : It doesn't compensate. It compensates for these values which the limit exists and the sign of the infinity on the exponent is defined by the strict sign of $n+1$ (not equivalent to $0$).
$endgroup$
– Rebellos
Dec 22 '18 at 0:00
$begingroup$
Presumably W. Alpha considers $0cdot infty$ to be $0.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:38
$begingroup$
@DanielWainfleet: No, you cannot assume WA actually does it right; see my comment on the question.
$endgroup$
– user21820
Dec 28 '18 at 4:01
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
It probably denotes that the final value of the limit depends on $n$. Specifically, if $n+1 < 0$, then the limit is $to e^{-infty} = 0$ while if $n+1 >0$, then the limit is $to e^{+infty} = infty$.
answered Dec 21 '18 at 23:58
RebellosRebellos
15.7k31250
$endgroup$
$begingroup$
But for certain values of $n$ if the limit is not defined then how does it compensate for that?
$endgroup$
– Shrey Joshi
Dec 21 '18 at 23:59
$begingroup$
And if $n+1=0$ it is $limlimits_{x rightarrow 0}{x ,{cot{x}}} = e^0=1$
$endgroup$
– Henry
Dec 22 '18 at 0:00
$begingroup$
Simple answer : It doesn't compensate. It compensates for these values which the limit exists and the sign of the infinity on the exponent is defined by the strict sign of $n+1$ (not equivalent to $0$).
$endgroup$
– Rebellos
Dec 22 '18 at 0:00
$begingroup$
Presumably W. Alpha considers $0cdot infty$ to be $0.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:38
$begingroup$
@DanielWainfleet: No, you cannot assume WA actually does it right; see my comment on the question.
$endgroup$
– user21820
Dec 28 '18 at 4:01
|
show 2 more comments
2
$begingroup$
It probably denotes that the final value of the limit depends on $n$. Specifically, if $n+1 < 0$, then the limit is $to e^{-infty} = 0$ while if $n+1 >0$, then the limit is $to e^{+infty} = infty$.
answered Dec 21 '18 at 23:58
RebellosRebellos
15.7k31250
$endgroup$
$begingroup$
But for certain values of $n$ if the limit is not defined then how does it compensate for that?
$endgroup$
– Shrey Joshi
Dec 21 '18 at 23:59
$begingroup$
And if $n+1=0$ it is $limlimits_{x rightarrow 0}{x ,{cot{x}}} = e^0=1$
$endgroup$
– Henry
Dec 22 '18 at 0:00
$begingroup$
Simple answer : It doesn't compensate. It compensates for these values which the limit exists and the sign of the infinity on the exponent is defined by the strict sign of $n+1$ (not equivalent to $0$).
$endgroup$
– Rebellos
Dec 22 '18 at 0:00
$begingroup$
Presumably W. Alpha considers $0cdot infty$ to be $0.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:38
$begingroup$
@DanielWainfleet: No, you cannot assume WA actually does it right; see my comment on the question.
$endgroup$
– user21820
Dec 28 '18 at 4:01
|
show 2 more comments
2
2
2
$begingroup$
It probably denotes that the final value of the limit depends on $n$. Specifically, if $n+1 < 0$, then the limit is $to e^{-infty} = 0$ while if $n+1 >0$, then the limit is $to e^{+infty} = infty$.
answered Dec 21 '18 at 23:58
RebellosRebellos
15.7k31250
$endgroup$
It probably denotes that the final value of the limit depends on $n$. Specifically, if $n+1 < 0$, then the limit is $to e^{-infty} = 0$ while if $n+1 >0$, then the limit is $to e^{+infty} = infty$.
answered Dec 21 '18 at 23:58
RebellosRebellos
15.7k31250
answered Dec 21 '18 at 23:58
RebellosRebellos
15.7k31250
answered Dec 21 '18 at 23:58
RebellosRebellos
15.7k31250
answered Dec 21 '18 at 23:58
RebellosRebellos
15.7k31250
15.7k31250
$begingroup$
But for certain values of $n$ if the limit is not defined then how does it compensate for that?
$endgroup$
– Shrey Joshi
Dec 21 '18 at 23:59
$begingroup$
And if $n+1=0$ it is $limlimits_{x rightarrow 0}{x ,{cot{x}}} = e^0=1$
$endgroup$
– Henry
Dec 22 '18 at 0:00
$begingroup$
Simple answer : It doesn't compensate. It compensates for these values which the limit exists and the sign of the infinity on the exponent is defined by the strict sign of $n+1$ (not equivalent to $0$).
$endgroup$
– Rebellos
Dec 22 '18 at 0:00
$begingroup$
Presumably W. Alpha considers $0cdot infty$ to be $0.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:38
$begingroup$
@DanielWainfleet: No, you cannot assume WA actually does it right; see my comment on the question.
$endgroup$
– user21820
Dec 28 '18 at 4:01
|
show 2 more comments
$begingroup$
But for certain values of $n$ if the limit is not defined then how does it compensate for that?
$endgroup$
– Shrey Joshi
Dec 21 '18 at 23:59
$begingroup$
And if $n+1=0$ it is $limlimits_{x rightarrow 0}{x ,{cot{x}}} = e^0=1$
$endgroup$
– Henry
Dec 22 '18 at 0:00
$begingroup$
Simple answer : It doesn't compensate. It compensates for these values which the limit exists and the sign of the infinity on the exponent is defined by the strict sign of $n+1$ (not equivalent to $0$).
$endgroup$
– Rebellos
Dec 22 '18 at 0:00
$begingroup$
Presumably W. Alpha considers $0cdot infty$ to be $0.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:38
$begingroup$
@DanielWainfleet: No, you cannot assume WA actually does it right; see my comment on the question.
$endgroup$
– user21820
Dec 28 '18 at 4:01
$begingroup$
But for certain values of $n$ if the limit is not defined then how does it compensate for that?
$endgroup$
– Shrey Joshi
Dec 21 '18 at 23:59
$begingroup$
But for certain values of $n$ if the limit is not defined then how does it compensate for that?
$endgroup$
– Shrey Joshi
Dec 21 '18 at 23:59
$begingroup$
And if $n+1=0$ it is $limlimits_{x rightarrow 0}{x ,{cot{x}}} = e^0=1$
$endgroup$
– Henry
Dec 22 '18 at 0:00
$begingroup$
And if $n+1=0$ it is $limlimits_{x rightarrow 0}{x ,{cot{x}}} = e^0=1$
$endgroup$
– Henry
Dec 22 '18 at 0:00
$begingroup$
Simple answer : It doesn't compensate. It compensates for these values which the limit exists and the sign of the infinity on the exponent is defined by the strict sign of $n+1$ (not equivalent to $0$).
$endgroup$
– Rebellos
Dec 22 '18 at 0:00
$begingroup$
Simple answer : It doesn't compensate. It compensates for these values which the limit exists and the sign of the infinity on the exponent is defined by the strict sign of $n+1$ (not equivalent to $0$).
$endgroup$
– Rebellos
Dec 22 '18 at 0:00
$begingroup$
Presumably W. Alpha considers $0cdot infty$ to be $0.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:38
$begingroup$
Presumably W. Alpha considers $0cdot infty$ to be $0.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:38
$begingroup$
@DanielWainfleet: No, you cannot assume WA actually does it right; see my comment on the question.
$endgroup$
– user21820
Dec 28 '18 at 4:01
$begingroup$
@DanielWainfleet: No, you cannot assume WA actually does it right; see my comment on the question.
$endgroup$
– user21820
Dec 28 '18 at 4:01
|
show 2 more comments
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2
$begingroup$
I think if you write that as an answer on an exam, it will be marked wrong.
$endgroup$
– GEdgar
Dec 22 '18 at 1:06
$begingroup$
WA often gives Wrong Answers for limits, because its programmers anyhow use ad-hoc heuristics. I have experienced them doing that: Sometimes, the wrong answer would even appear before it is substituted by the correct answer. Other times, contradictory answers would appear. At any point in time (including in the future), you can easily come up with an infinite class of limits that WA will give the Wrong Answer to. I'll leave it as a fun exercise for anyone interested.
$endgroup$
– user21820
Dec 28 '18 at 4:00