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

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?

• Does $x$ go over the integers or the real numbers here? – Arthur Mar 11 '19 at 19:21
• There isn't any additional information provided. So, I'm assuming it is real numbers. – Eldar Rahimli Mar 11 '19 at 19:22
• $x$ 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) – Cardioid_Ass_22 Mar 11 '19 at 19:28
• "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. – fleablood Mar 11 '19 at 19:40
• 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. – Matthew Liu Mar 11 '19 at 20:03

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.
• 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? – Eldar Rahimli Mar 11 '19 at 19:32