Why doesn't this limit exist? I'm trying to compute this limit:
$\lim_{x\to +\infty}x^2-x^2\sqrt{1+\frac{(-1)^x}{x^2}}$
which, according to the results, is not defined; I initially thought about a $\infty - \infty$ form but it looks like I'm missing something...
 A: With $\;x\in\Bbb N\;$ , so that $\;(-1)^x\;$ has a definite meaning within the real numbers:
$$\left(x^2-x^2\sqrt{1+\frac{(-1)^x}{x^2}}\right)\cdot\frac{x^2+x^2\sqrt{1+\frac{(-1)^x}{x^2}}}{x^2+x^2\sqrt{1+\frac{(-1)^x}{x^2}}}=$$$${}$$
$$=\frac{(-1)^{x+1}x^2}{x^2+x^2\sqrt{1+\cfrac{(-1)^x}{x^2}}}\xrightarrow[x\to\infty]{}\begin{cases}\;\;\,\frac12,&x\;\text{is odd}\\{}\\-\frac12,&x\;\text{is even}\end{cases}$$
and thus the limit doesn't exist.
A: Note first that there's a potential issue with $(-1)^x$ being ambiguous or undefined for non-integer values of $x$. That's not a fundamental problem, however; we can just consider the question as asking about a sequence.
With that in mind, consider we can multiply by the conjugate root:
\begin{align*}
x^2 - x^2\sqrt{1 + \frac{(-1)^x}{x^2}}
&= \frac{x^4 - x^4\left(1 + \frac{(-1)^x}{x^2}\right)}{x^2 + x^2\sqrt{1 + \frac{(-1)^x}{x^2}}} \\
&= \frac{-x^2(-1)^x}{x^2 + x^2\sqrt{1 + \frac{(-1)^x}{x^2}}} \\
&= \frac{-(-1)^x}{1 + \sqrt{1 + \frac{(-1)^x}{x^2}}} \\
&= (-1)^{x+1} \frac{1}{1 + \sqrt{1+\frac{(-1)^x}{x^2}}}
\end{align*}
Notice that $\frac{1}{1 + \sqrt{1+\frac{(-1)^x}{x^2}}}$ is always positive and does not approach 0. Therefore, since the $(-1)^{x+1}$ term out front makes the whole expression alternate signs, the limit does not exist.
