$$ \lim\limits_{x\to2}(-1/2)^x=1/4$$
My teacher says this is not correct, and I just can't fathom why this is not correct. I tried to find the value of the given function in the neighbourhood of $2$, and I got this
Now I can't understand why my calculator is giving these weird results.
are $(-1/2)^{1.92}$ and $(-1/2)^{192/100}$ different
and does this limit really not exist?
Best Answer
Let us define the "real-valued" exponential $f$ as the partial function given by
$$ f(x) = [\text{unique real value among $(-1/2)^x = e^{x\log(-1/2)}$} ], $$
where $\log(z) = \log|z| + i(2\pi k + \arg(z))$ is the multivalued complex logarithm. Now there are several observations:
$f(x)$ can only possibly defined if $x$ is rational.
If $x = p/q$ is a rational number in lowest terms, then we can prove that the set $$\{e^{(p/q)\log(-1/2)} = (1/2)^{p/q}e^{(2k+1)\pi i p/q} : k \in \mathbb{Z}\}$$ contains exactly one real number if $q$ is odd, and no real numbers if $q$ is even. Consequently, the domain of $f$ is $$ \operatorname{dom}(f) = \{ p/q : \text{$p, q \in \mathbb{Z}$ s.t. $\gcd(p, q) = 1$ and $q$ is odd} \}. $$ This is a dense subset of $\mathbb{R}$, so the notion of limit of $f$ can be discussed everywhere on $\mathbb{R}$.
For each $p/q \in \operatorname{dom}(f)$ in lowest term, it turns out that $$ f(p/q) = (-1)^p (1/2)^{p/q}. $$ For example \begin{align*} f(1.92) &= f(48/25) = (-1)^{48}(1/2)^{48/25} \approx 0.264255, \\ f(1.96) &= f(49/25) = (-1)^{49}(1/2)^{49/25} \approx -0.257028. \end{align*} (Note: If the exponential function is implemented differently in your calculator, the outcome can certainly be different from what $f(t)$ evaluates.)
Under this definition, we can show that $f$ does not have limit at any point. Indeed, this is because both
$$ \{ x \in \operatorname{dom}(f) : f(x) > 0 \} = \{ p/q : \text{$p, q \in \mathbb{Z}$ s.t. $\gcd(p, q) = 1$ and $p$ is even and $q$ is odd} \} $$
and
$$ \{ x \in \operatorname{dom}(f) : f(x) < 0 \} = \{ p/q : \text{$p, q \in \mathbb{Z}$ s.t. $\gcd(p, q) = 1$ and both $p$ and $q$ are odd} \} $$
are dense subsets of $\mathbb{R}$. More precisely, for any $x \in \mathbb{R}$,
$$ \limsup_{\operatorname{dom}(f) \ni t \to x} f(t) = (1/2)^x \qquad \text{and}\qquad \liminf_{\operatorname{dom}(f) \ni t \to x} f(t) = -(1/2)^{x}. $$
Since limsup and liminf are different, the limit cannot exist. (Of course, the non-existence of limit can also be shown by finding subsequences that tend to different values.)
Remark. Note that the "real-valued" exponential $f$ is a "maximal" version. However, if we restrict $f$ further onto some dense subset $D$, then the restriction $f|_D$ can possibly have limits. (I suspect that the calculator OP is using implements the exponential in such a way that $(-1/2)^x$ is chosen to be real only on certain subsets of $\operatorname{dom}(f)$. In this case, the answer does depend on that particular implementation of the exponential function.)