Show using $\epsilon - \delta$ definition of limit that $\lim_{x\to 1} -x^4 = -1$ and find the value of $\delta$.
Note that\begin{align}|-x^4-(-1)|&=|x^4-1|\\&=|x-1|\times|x^3+x^2+x+1|.\end{align}And, if $|x-1|<1$, then$$|x|=|(x-1)+1|\leqslant|x-1|+1<2,$$and therefore$$|x^3+x^2+x+1|\leqslant2^3+2^2+2+1=15,$$which implies that $|-x^4-(-1)|<15|x-1|$. So, if $|x-1|<\frac\varepsilon{15}$, then $|-x^4-(-1)|<\varepsilon$. In particular, $\delta=\min\left\{1,\frac\varepsilon{15}\right\}$ will work.