Evaluate $lim_xto infty(-3)^2x+1$?How do I evaluate $lim_h to infty e^h(1-s)$?A Gamma limit $lim_nrightarrow+inftysum_k=1^n left( Gammabigl(fracknbigr)right)^-k=frace^gammae^gamma-1$Evaluate the following limit of finite sumdividing expressions with limitsEuler Limit with -1 to infinity?Formal way to evaluate this limitEvaluate $lim_x to -infty ln(-x^3+x)$Limit of composite functionWhy does the following not show $zeta(0) = -frac12$?limit of Integral of product of convergent functions
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Evaluate $lim_xto infty(-3)^2x+1$?
How do I evaluate $lim_h to infty e^h(1-s)$?A Gamma limit $lim_nrightarrow+inftysum_k=1^n left( Gammabigl(fracknbigr)right)^-k=frace^gammae^gamma-1$Evaluate the following limit of finite sumdividing expressions with limitsEuler Limit with -1 to infinity?Formal way to evaluate this limitEvaluate $lim_x to -infty ln(-x^3+x)$Limit of composite functionWhy does the following not show $zeta(0) = -frac12$?limit of Integral of product of convergent functions
$begingroup$
Evaluate $lim_xto infty(-3)^2x+1?$
I have two solutions for this problem and both of them look valid to me, even though one of them is incorrect.
Firstly, I know that $(2x+1)$ is an odd number, so the limit will be positive infinity when $x$ takes an even number, and limit will be negative infinity when $x$ takes an odd number. Therefore, the limit does not exist.
On the other hand, if I perform some mathematical operations on the function:
$$lim_xto infty(-3)^2x+1=lim_xto inftybigl((-3)^2bigr)^xcdot(-3)=lim_xto infty9^xcdot(-3)=-9^infty=-infty$$
Where am I making a mistake on my second attempt?
calculus limits
asked Mar 11 at 19:19
Eldar RahimliEldar Rahimli
35110
$endgroup$
|
show 7 more comments
$begingroup$
Evaluate $lim_xto infty(-3)^2x+1?$
I have two solutions for this problem and both of them look valid to me, even though one of them is incorrect.
Firstly, I know that $(2x+1)$ is an odd number, so the limit will be positive infinity when $x$ takes an even number, and limit will be negative infinity when $x$ takes an odd number. Therefore, the limit does not exist.
On the other hand, if I perform some mathematical operations on the function:
$$lim_xto infty(-3)^2x+1=lim_xto inftybigl((-3)^2bigr)^xcdot(-3)=lim_xto infty9^xcdot(-3)=-9^infty=-infty$$
Where am I making a mistake on my second attempt?
calculus limits
asked Mar 11 at 19:19
Eldar RahimliEldar Rahimli
35110
$endgroup$
2
$begingroup$
Does $x$ go over the integers or the real numbers here?
$endgroup$
– Arthur
Mar 11 at 19:21
$begingroup$
There isn't any additional information provided. So, I'm assuming it is real numbers.
$endgroup$
– Eldar Rahimli
Mar 11 at 19:22
1
$begingroup$
$x$ being even or odd doesn't matter- as long as $x$ is an integer, $2x+1$ will always be an odd integer (regardless of the parity of $x$ itself)
$endgroup$
– Cardioid_Ass_22
Mar 11 at 19:28
1
$begingroup$
"There isn't any additional information provided. So, I'm assuming it is real numbers. " Limit doesn't exist for real numbers. If $2x+1$ is an even integer the value is positive and if $2x+1$ is odd integer it is negative. If $2x + 1 = frac nm$ where $m$ is even it is not defined. It is not defined if $2x + 1$ is irrational.
$endgroup$
– fleablood
Mar 11 at 19:40
1
$begingroup$
You cannot perform nonsense mathematical manipulations if you have no idea what you are doing. $-3^2x$ is most definitely NOT $9^x$ for many real values of x.
$endgroup$
– Matthew Liu
Mar 11 at 20:03
|
show 7 more comments
$begingroup$
Evaluate $lim_xto infty(-3)^2x+1?$
I have two solutions for this problem and both of them look valid to me, even though one of them is incorrect.
Firstly, I know that $(2x+1)$ is an odd number, so the limit will be positive infinity when $x$ takes an even number, and limit will be negative infinity when $x$ takes an odd number. Therefore, the limit does not exist.
On the other hand, if I perform some mathematical operations on the function:
$$lim_xto infty(-3)^2x+1=lim_xto inftybigl((-3)^2bigr)^xcdot(-3)=lim_xto infty9^xcdot(-3)=-9^infty=-infty$$
Where am I making a mistake on my second attempt?
calculus limits
asked Mar 11 at 19:19
Eldar RahimliEldar Rahimli
35110
$endgroup$
Evaluate $lim_xto infty(-3)^2x+1?$
I have two solutions for this problem and both of them look valid to me, even though one of them is incorrect.
Firstly, I know that $(2x+1)$ is an odd number, so the limit will be positive infinity when $x$ takes an even number, and limit will be negative infinity when $x$ takes an odd number. Therefore, the limit does not exist.
On the other hand, if I perform some mathematical operations on the function:
$$lim_xto infty(-3)^2x+1=lim_xto inftybigl((-3)^2bigr)^xcdot(-3)=lim_xto infty9^xcdot(-3)=-9^infty=-infty$$
Where am I making a mistake on my second attempt?
calculus limits
calculus limits
asked Mar 11 at 19:19
Eldar RahimliEldar Rahimli
35110
asked Mar 11 at 19:19
Eldar RahimliEldar Rahimli
35110
edited Mar 11 at 20:07
Eldar Rahimli
asked Mar 11 at 19:19
Eldar RahimliEldar Rahimli
35110
asked Mar 11 at 19:19
Eldar RahimliEldar Rahimli
35110
asked Mar 11 at 19:19
Eldar RahimliEldar Rahimli
35110
35110
2
$begingroup$
Does $x$ go over the integers or the real numbers here?
$endgroup$
– Arthur
Mar 11 at 19:21
$begingroup$
There isn't any additional information provided. So, I'm assuming it is real numbers.
$endgroup$
– Eldar Rahimli
Mar 11 at 19:22
1
$begingroup$
$x$ being even or odd doesn't matter- as long as $x$ is an integer, $2x+1$ will always be an odd integer (regardless of the parity of $x$ itself)
$endgroup$
– Cardioid_Ass_22
Mar 11 at 19:28
1
$begingroup$
"There isn't any additional information provided. So, I'm assuming it is real numbers. " Limit doesn't exist for real numbers. If $2x+1$ is an even integer the value is positive and if $2x+1$ is odd integer it is negative. If $2x + 1 = frac nm$ where $m$ is even it is not defined. It is not defined if $2x + 1$ is irrational.
$endgroup$
– fleablood
Mar 11 at 19:40
1
$begingroup$
You cannot perform nonsense mathematical manipulations if you have no idea what you are doing. $-3^2x$ is most definitely NOT $9^x$ for many real values of x.
$endgroup$
– Matthew Liu
Mar 11 at 20:03
|
show 7 more comments
2
$begingroup$
Does $x$ go over the integers or the real numbers here?
$endgroup$
– Arthur
Mar 11 at 19:21
$begingroup$
There isn't any additional information provided. So, I'm assuming it is real numbers.
$endgroup$
– Eldar Rahimli
Mar 11 at 19:22
1
$begingroup$
$x$ being even or odd doesn't matter- as long as $x$ is an integer, $2x+1$ will always be an odd integer (regardless of the parity of $x$ itself)
$endgroup$
– Cardioid_Ass_22
Mar 11 at 19:28
1
$begingroup$
"There isn't any additional information provided. So, I'm assuming it is real numbers. " Limit doesn't exist for real numbers. If $2x+1$ is an even integer the value is positive and if $2x+1$ is odd integer it is negative. If $2x + 1 = frac nm$ where $m$ is even it is not defined. It is not defined if $2x + 1$ is irrational.
$endgroup$
– fleablood
Mar 11 at 19:40
1
$begingroup$
You cannot perform nonsense mathematical manipulations if you have no idea what you are doing. $-3^2x$ is most definitely NOT $9^x$ for many real values of x.
$endgroup$
– Matthew Liu
Mar 11 at 20:03
2
2
$begingroup$
Does $x$ go over the integers or the real numbers here?
$endgroup$
– Arthur
Mar 11 at 19:21
$begingroup$
Does $x$ go over the integers or the real numbers here?
$endgroup$
– Arthur
Mar 11 at 19:21
$begingroup$
There isn't any additional information provided. So, I'm assuming it is real numbers.
$endgroup$
– Eldar Rahimli
Mar 11 at 19:22
$begingroup$
There isn't any additional information provided. So, I'm assuming it is real numbers.
$endgroup$
– Eldar Rahimli
Mar 11 at 19:22
1
1
$begingroup$
$x$ being even or odd doesn't matter- as long as $x$ is an integer, $2x+1$ will always be an odd integer (regardless of the parity of $x$ itself)
$endgroup$
– Cardioid_Ass_22
Mar 11 at 19:28
$begingroup$
$x$ being even or odd doesn't matter- as long as $x$ is an integer, $2x+1$ will always be an odd integer (regardless of the parity of $x$ itself)
$endgroup$
– Cardioid_Ass_22
Mar 11 at 19:28
1
1
$begingroup$
"There isn't any additional information provided. So, I'm assuming it is real numbers. " Limit doesn't exist for real numbers. If $2x+1$ is an even integer the value is positive and if $2x+1$ is odd integer it is negative. If $2x + 1 = frac nm$ where $m$ is even it is not defined. It is not defined if $2x + 1$ is irrational.
$endgroup$
– fleablood
Mar 11 at 19:40
$begingroup$
"There isn't any additional information provided. So, I'm assuming it is real numbers. " Limit doesn't exist for real numbers. If $2x+1$ is an even integer the value is positive and if $2x+1$ is odd integer it is negative. If $2x + 1 = frac nm$ where $m$ is even it is not defined. It is not defined if $2x + 1$ is irrational.
$endgroup$
– fleablood
Mar 11 at 19:40
1
1
$begingroup$
You cannot perform nonsense mathematical manipulations if you have no idea what you are doing. $-3^2x$ is most definitely NOT $9^x$ for many real values of x.
$endgroup$
– Matthew Liu
Mar 11 at 20:03
$begingroup$
You cannot perform nonsense mathematical manipulations if you have no idea what you are doing. $-3^2x$ is most definitely NOT $9^x$ for many real values of x.
$endgroup$
– Matthew Liu
Mar 11 at 20:03
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
A negative number raised to an odd number is always negative, so "change between negative and positive" is wrong.
So assuming $x$ goes over the integers, the limit is $-infty$ as the exponent $2x+1$ is always odd. Yes, $(-3)^x$ alternates between positive and negative, but $(-3)^2x+1$ is strictly negative.
If $x$ goes over the reals, then before you can evaluate the limit you have to decide what something like $(-3)^pi$ means.
answered Mar 11 at 19:24
ArthurArthur
118k7117200
$endgroup$
$begingroup$
I've edited the question and clarified what I meant by "change between negative and positive". So, can we say that the problem should have stated $xin Z$ to be solvable?
$endgroup$
– Eldar Rahimli
Mar 11 at 19:32
$begingroup$
@EldarRahimli Yes, I think it really should have done that.
$endgroup$
– Arthur
Mar 11 at 19:36
add a comment |
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$begingroup$
A negative number raised to an odd number is always negative, so "change between negative and positive" is wrong.
So assuming $x$ goes over the integers, the limit is $-infty$ as the exponent $2x+1$ is always odd. Yes, $(-3)^x$ alternates between positive and negative, but $(-3)^2x+1$ is strictly negative.
If $x$ goes over the reals, then before you can evaluate the limit you have to decide what something like $(-3)^pi$ means.
answered Mar 11 at 19:24
ArthurArthur
118k7117200
$endgroup$
$begingroup$
I've edited the question and clarified what I meant by "change between negative and positive". So, can we say that the problem should have stated $xin Z$ to be solvable?
$endgroup$
– Eldar Rahimli
Mar 11 at 19:32
$begingroup$
@EldarRahimli Yes, I think it really should have done that.
$endgroup$
– Arthur
Mar 11 at 19:36
add a comment |
$begingroup$
A negative number raised to an odd number is always negative, so "change between negative and positive" is wrong.
So assuming $x$ goes over the integers, the limit is $-infty$ as the exponent $2x+1$ is always odd. Yes, $(-3)^x$ alternates between positive and negative, but $(-3)^2x+1$ is strictly negative.
If $x$ goes over the reals, then before you can evaluate the limit you have to decide what something like $(-3)^pi$ means.
answered Mar 11 at 19:24
ArthurArthur
118k7117200
$endgroup$
$begingroup$
I've edited the question and clarified what I meant by "change between negative and positive". So, can we say that the problem should have stated $xin Z$ to be solvable?
$endgroup$
– Eldar Rahimli
Mar 11 at 19:32
$begingroup$
@EldarRahimli Yes, I think it really should have done that.
$endgroup$
– Arthur
Mar 11 at 19:36
add a comment |
$begingroup$
A negative number raised to an odd number is always negative, so "change between negative and positive" is wrong.
So assuming $x$ goes over the integers, the limit is $-infty$ as the exponent $2x+1$ is always odd. Yes, $(-3)^x$ alternates between positive and negative, but $(-3)^2x+1$ is strictly negative.
If $x$ goes over the reals, then before you can evaluate the limit you have to decide what something like $(-3)^pi$ means.
answered Mar 11 at 19:24
ArthurArthur
118k7117200
$endgroup$
A negative number raised to an odd number is always negative, so "change between negative and positive" is wrong.
So assuming $x$ goes over the integers, the limit is $-infty$ as the exponent $2x+1$ is always odd. Yes, $(-3)^x$ alternates between positive and negative, but $(-3)^2x+1$ is strictly negative.
If $x$ goes over the reals, then before you can evaluate the limit you have to decide what something like $(-3)^pi$ means.
answered Mar 11 at 19:24
ArthurArthur
118k7117200
edited Mar 11 at 19:35
answered Mar 11 at 19:24
ArthurArthur
118k7117200
answered Mar 11 at 19:24
ArthurArthur
118k7117200
answered Mar 11 at 19:24
ArthurArthur
118k7117200
118k7117200
$begingroup$
I've edited the question and clarified what I meant by "change between negative and positive". So, can we say that the problem should have stated $xin Z$ to be solvable?
$endgroup$
– Eldar Rahimli
Mar 11 at 19:32
$begingroup$
@EldarRahimli Yes, I think it really should have done that.
$endgroup$
– Arthur
Mar 11 at 19:36
add a comment |
$begingroup$
I've edited the question and clarified what I meant by "change between negative and positive". So, can we say that the problem should have stated $xin Z$ to be solvable?
$endgroup$
– Eldar Rahimli
Mar 11 at 19:32
$begingroup$
@EldarRahimli Yes, I think it really should have done that.
$endgroup$
– Arthur
Mar 11 at 19:36
$begingroup$
I've edited the question and clarified what I meant by "change between negative and positive". So, can we say that the problem should have stated $xin Z$ to be solvable?
$endgroup$
– Eldar Rahimli
Mar 11 at 19:32
$begingroup$
I've edited the question and clarified what I meant by "change between negative and positive". So, can we say that the problem should have stated $xin Z$ to be solvable?
$endgroup$
– Eldar Rahimli
Mar 11 at 19:32
$begingroup$
@EldarRahimli Yes, I think it really should have done that.
$endgroup$
– Arthur
Mar 11 at 19:36
$begingroup$
@EldarRahimli Yes, I think it really should have done that.
$endgroup$
– Arthur
Mar 11 at 19:36
add a comment |
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2
$begingroup$
Does $x$ go over the integers or the real numbers here?
$endgroup$
– Arthur
Mar 11 at 19:21
$begingroup$
There isn't any additional information provided. So, I'm assuming it is real numbers.
$endgroup$
– Eldar Rahimli
Mar 11 at 19:22
1
$begingroup$
$x$ being even or odd doesn't matter- as long as $x$ is an integer, $2x+1$ will always be an odd integer (regardless of the parity of $x$ itself)
$endgroup$
– Cardioid_Ass_22
Mar 11 at 19:28
1
$begingroup$
"There isn't any additional information provided. So, I'm assuming it is real numbers. " Limit doesn't exist for real numbers. If $2x+1$ is an even integer the value is positive and if $2x+1$ is odd integer it is negative. If $2x + 1 = frac nm$ where $m$ is even it is not defined. It is not defined if $2x + 1$ is irrational.
$endgroup$
– fleablood
Mar 11 at 19:40
1
$begingroup$
You cannot perform nonsense mathematical manipulations if you have no idea what you are doing. $-3^2x$ is most definitely NOT $9^x$ for many real values of x.
$endgroup$
– Matthew Liu
Mar 11 at 20:03