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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 $lim_{nrightarrow+infty}sum_{k=1}^n left( Gammabigl(frac{k}{n}bigr)right)^{-k}=frac{e^gamma}{e^gamma-1}$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 $zeta(0) = -frac{1}{2}$?limit of Integral of product of convergent functions

0

$begingroup$

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:

$$lim_{xto infty}{(-3)^{2x+1}}=lim_{xto infty}{bigl((-3)^2bigr)^xcdot(-3)}=lim_{xto infty}{9^xcdot(-3)}=-9^infty=-infty$$

Where am I making a mistake on my second attempt?

calculus limits

share|cite|improve this question

edited Mar 11 at 20:07

Eldar Rahimli

asked Mar 11 at 19:19

Eldar RahimliEldar Rahimli

36310

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  Does $x$ go over the integers or the real numbers here?
  $endgroup$
  – Arthur
  Mar 11 at 19:21
  

  
  
  
  


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  There isn't any additional information provided. So, I'm assuming it is real numbers.
  $endgroup$
  – Eldar Rahimli
  Mar 11 at 19:22
  

  
  
  
  


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  $x$ 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
  

  
  
  
  


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  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
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
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  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

0

$begingroup$

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:

$$lim_{xto infty}{(-3)^{2x+1}}=lim_{xto infty}{bigl((-3)^2bigr)^xcdot(-3)}=lim_{xto infty}{9^xcdot(-3)}=-9^infty=-infty$$

Where am I making a mistake on my second attempt?

calculus limits

share|cite|improve this question

edited Mar 11 at 20:07

Eldar Rahimli

asked Mar 11 at 19:19

Eldar RahimliEldar Rahimli

36310

$endgroup$

• 
  
  
  

  
  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:

$$lim_{xto infty}{(-3)^{2x+1}}=lim_{xto infty}{bigl((-3)^2bigr)^xcdot(-3)}=lim_{xto infty}{9^xcdot(-3)}=-9^infty=-infty$$

Where am I making a mistake on my second attempt?

calculus limits

share|cite|improve this question

edited Mar 11 at 20:07

Eldar Rahimli

asked Mar 11 at 19:19

Eldar RahimliEldar Rahimli

36310

$endgroup$

share|cite|improve this question

edited Mar 11 at 20:07

Eldar Rahimli

asked Mar 11 at 19:19

Eldar RahimliEldar Rahimli

36310

share|cite|improve this question

edited Mar 11 at 20:07

Eldar Rahimli

asked Mar 11 at 19:19

Eldar RahimliEldar Rahimli

36310

asked Mar 11 at 19:19

Eldar RahimliEldar Rahimli

36310

asked Mar 11 at 19:19

Eldar RahimliEldar Rahimli

36310

1 Answer
1

active

oldest

votes

3

$begingroup$

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.

share|cite|improve this answer

edited Mar 11 at 19:35

answered Mar 11 at 19:24

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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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3

$begingroup$

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.

share|cite|improve this answer

edited Mar 11 at 19:35

answered Mar 11 at 19:24

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

3

$begingroup$

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.

share|cite|improve this answer

edited Mar 11 at 19:35

answered Mar 11 at 19:24

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.

share|cite|improve this answer

edited Mar 11 at 19:35

answered Mar 11 at 19:24

ArthurArthur

118k7118201

$endgroup$

share|cite|improve this answer

edited Mar 11 at 19:35

answered Mar 11 at 19:24

ArthurArthur

118k7118201

answered Mar 11 at 19:24

ArthurArthur

118k7118201

answered Mar 11 at 19:24

ArthurArthur

118k7118201