Answer
There are 3 points $c$ that we can automatically know $\lim_{x\to c}f(x)$: $c=0$, $c=1$ and $c=-1$.
$\lim_{x\to0}f(x)=0$
$\lim_{x\to1}f(x)=\lim_{x\to-1}f(x)=1$
Work Step by Step
We can see here that whichever $x$ in $R$, $f(x)$ is still squeezed between $x^4$ and $x^2$. So the Sandwich Theorem can be applied in $R$.
To automatically know $\lim_{x\to c}f(x)$, we need to rely on the Sandwich Theorem, which would let us know $\lim_{x\to c}f(x)$ if we already know $\lim_{x\to c}(x^4)$ and $\lim_{x\to c}(x^2)$ with the condition that $\lim_{x\to c}(x^4)=\lim_{x\to c}(x^2)$
In short, at which $c$ in $R$ can we know immediately that $\lim_{x\to c}(x^4)=\lim_{x\to c}(x^2)$, or in fact, $$c^4=c^2$$ $$c^4-c^2=0$$ $$c^2(c^2-1)=0$$ $$c^2=0\hspace{1cm}\text{or}\hspace{1cm}c^2=1$$ $$c=0\hspace{1cm}\text{or}\hspace{1cm}c=-1\hspace{1cm}\text{or}\hspace{1cm}c=1$$
- At $c=0$: $\lim_{x\to0}(x^2)=\lim_{x\to0}(x^4)=0^2=0$. Therefore, $\lim_{x\to0}f(x)=0$
- At $c=1$: $\lim_{x\to1}(x^2)=\lim_{x\to0}(x^4)=1^2=1$. Therefore, $\lim_{x\to1}f(x)=1$
- At $c=-1$: $\lim_{x\to-1}(x^2)=\lim_{x\to-1}(x^4)=(-1)^2=1$. Therefore, $\lim_{x\to-1}f(x)=1$
