Composition of Piecewise Function Limit

Note that $x^2 > 0$ for $x \neq 0$, so if set $1-x^2=t$, we have $t < 1$.

So we have that $$\lim_{x \to 0} f(1-x^2)=\lim_{t \to 1^{-}}f(t)=\lim_{t \to 1^{-}}(t^2+2)=3$$ As $f(t)=t^2+2$ for $t<1$.

$$\lim_{x \to 0}f(1-x^2)=\lim_{x\to0}(1-x^2)^2+2=(1-0)^2+2=3$$ Because $x\to0$, we can use the $x<1$ function and just plug in.

EDIT: The above statement implies, but we must explicitly say that for all $x\not=0$, $x^2>0$ therefore, $1-x^2<1$

The limit of a function at a given point is determined by the values it takes near, but not at, that point. For example, to determine the limit of $x \mapsto f(1 - x^2)$ at zero, we need to examine the values of $f(1 - x^2)$ at values of $x$ that are close to but not equal to zero.

Assuming that $x$ is meant to be restricted to real numbers (which the piecewise definition of $f$ implicitly suggests), $x^2 > 0$ (and therefore $1 - x^2 < 1$) whenever $x \ne 0$. Thus: $$f(1 - x^2) = (1 - x^2)^2 + 2 \quad \forall x \ne 0.$$

The limit of $(1 - x^2)^2 + 2$ as $x \to 0$ equals $3$.