Transform the interval along with the series argument

The standard logarithm series represents $\ln(1+u)$ for $-1<u\le1$. With $u=x^2/2$, the required condition is $-1<x^2/2\le1$. The lower inequality holds automatically because $x^2/2\ge0$.

$x^2/2\le1$ gives $x^2\le2$, or $(x-\sqrt2)(x+\sqrt2)\le0$. The Teacher’s small upward-parabola sketch shows the nonpositive part between its two roots. Therefore $-\sqrt2\le x\le\sqrt2$.

At either endpoint $x=\pm\sqrt2$, the substituted argument is 1, which is included in the original logarithm-series interval. Both endpoints therefore remain included. Independently, the function $\ln(2+x^2)$ exists for every real x, but its Maclaurin representation here is limited to this closed interval; that distinction does not restrict the underlying function.