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Evaluate $\lim_{x\to \infty}{(-3)^{2x+1}}?$

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:

$$\lim_{x\to \infty}{(-3)^{2x+1}}=\lim_{x\to \infty}{\bigl((-3)^2\bigr)^x\cdot(-3)}=\lim_{x\to \infty}{9^x\cdot(-3)}=-9^\infty=-\infty$$

Where am I making a mistake on my second attempt?

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    $\begingroup$ Does $x$ go over the integers or the real numbers here? $\endgroup$ – Arthur Mar 11 '19 at 19:21
  • $\begingroup$ There isn't any additional information provided. So, I'm assuming it is real numbers. $\endgroup$ – Eldar Rahimli Mar 11 '19 at 19:22
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    $\begingroup$ $x$ 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 '19 at 19:28
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    $\begingroup$ "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 '19 at 19:40
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    $\begingroup$ 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 '19 at 20:03
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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.

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  • $\begingroup$ 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 $x\in Z$ to be solvable? $\endgroup$ – Eldar Rahimli Mar 11 '19 at 19:32
  • $\begingroup$ @EldarRahimli Yes, I think it really should have done that. $\endgroup$ – Arthur Mar 11 '19 at 19:36

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