If $\lim_{x \to \infty} (-1)^x$ is undefined ...

Why is $\lim_{x \to \infty} (-1/4)^x$ zero?

Couldn't you take out the negative to make it $\lim_{x \to \infty} (-1)^x$ * $\lim_{x \to \infty} (-1/4)^x$ which would make it undefined?

Does undefined * 0 = undefined or 0?

  • 5
    $\begingroup$ Good question: the answer is no. In fact, $\lim_{x \to \infty} f(x)g(x) = \lim_{x \to \infty} f(x) \lim_{x \to \infty} g(x)$ only if $\lim_{x \to \infty} f(x)$ and $\lim_{x \to \infty} g(x)$ both exist and are defined. In this case, $f(x) = (-1)^{x}, g(x) = (1/4)^{x}$. $\endgroup$ – Alex Wertheim Oct 12 '15 at 22:30

You have the following: let $x\in\mathbb{N}$ then the sequence $a_x$ $$ a_x:=(-1)^x $$ can only take two possible values, $a_{2x}=1, a_{2x+1}=-1$ and therefore the limit does not exist.

If you now look to the sequence $b_x$ with $$ b_x:=(-\frac14)^x $$ then you can make the following inequality work $$ -(\frac14)^x\leq (-\frac14)^x\leq(\frac14)^x $$ and $-(\frac14)^x, (\frac14)^x$ both tend to zero as $x$ goes to infinity. and therefore also $b_x=(-\frac14)^x $ goes to zero.

Now why can you not take out the $(-1)^x$ out of $b_x$? This is because the limit is not defined. if you want to $$ \lim_{x\to\infty}a_xb_x=\lim_{x\to\infty}a_x\lim_{x\to\infty}b_x $$ both limits on the RHS have to exist.


An easier way to look at the reason why $\lim_{x \to \infty} \left(\frac{-1}{4}\right)^x = 0$ is by continuously dividing $1$ by a really large number so that the answer would be closer and closer to zero.

$$\lim_{x\to \infty} \left(\frac{-1}{4}\right)^x \approx \lim_{x \to \infty}\frac{\pm 1}{4^x}$$

$$\begin{align} & \frac{1}{4^2} = \frac{1}{16} = 0.0625 \\ & \frac{1}{4^3} = \frac{1}{64} = 0.015625 \\ & \frac{1}{4^4} = \frac{1}{256} = 0.00390625 \\ \end{align}$$

So obviously as $x$ increases $\frac{1}{4^x}$ decreases, and since $x \to \infty$ you would obviously have a value of something near zero for very large values of $x$.

$$\frac{1}{4^{100}} = \frac{1}{160\, \cdots \,5301376} = 6.22 \cdot 10^{-61} \approx 0$$

As for the splitting part, you have to make sure that the limits would not be undefined so that they would be equal:

$$\begin{align} \lim_{x \to \infty} (f(x)g(x)) & = \lim_{x \to \infty} f(x) \lim_{x \to \infty} g(x) \\ & \ne (\text{undefined}) \cdot \lim_{x \to \infty} g(x) \\ & \ne \text{undefined} \\ \end{align}$$

In this case we would have:

$$\begin{align} \lim_{x \to \infty} \left(\frac{-1}{4}\right)^x & \ne \lim_{x \to \infty} \left((-1)^x \cdot \left(\frac{-1}{4}\right)^x\right)\\ & \ne (\text{undefined}) \cdot \lim_{x \to \infty} \left(\frac{-1}{4}\right)^x \\ & \ne \text{undefined} \\ \end{align}$$

Therefore factoring the $-1$ out of the fraction would be problematic, so I suggest that you just solve the limit directly as I have shown earlier.


Good question.

Notice that you are trying to make use of the rule that if $f$ and $g$ are two functions such that the limits $\lim_{x\to\infty} f(x)$ and $\lim_{x\to\infty} g(x)$ exist, then

$$\lim_{x\to\infty} f(x)g(x) = \lim_{x\to\infty} f(x)\lim_{x\to\infty} g(x).$$

That is, you want to split the limit of a product into the product of the limits - but in your case, we have $f(x) = (-1)^x$, and $g(x) = (\frac{1}{4})^x$, and it was necessary for the rule to apply that both limits $\lim_{x\to\infty} f(x)$ and $\lim_{x\to\infty} g(x)$ existed, but the limit for $f$ doesn't in this case.

In conclusion, the problem is that

$$\lim_{x\to\infty} \left(-\frac{1}{4}\right)^x \neq \lim_{x\to\infty}(-1)^x\lim_{x\to\infty}\left(\frac{1}{4}\right)^x,$$

where the expression on the right hand side doesn't even make sense.


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