Right-hand limit of f(x) as x approaches -1

When $x > -1$, we have $x + 1 > 0$, so $|x+1| = x+1$. Substituting into $f(x)$ gives $f(x) = \dfrac{x+1}{x+1} = 1$ for all such $x$.

Since $f(x)$ equals the constant value $1$ for every $x$ slightly greater than $-1$, the values in the table stay fixed at $1$ and do not change as $x \to -1^+$. Therefore $\displaystyle\lim_{x \to -1^+} f(x) = 1$.

When $x < -1$, we have $x + 1 < 0$, so $|x+1| = -(x+1)$. Substituting into $f(x)$ gives $f(x) = \dfrac{-(x+1)}{x+1} = -1$ for all such $x$.

Since $f(x)$ equals the constant value $-1$ for every $x$ slightly less than $-1$, the values in the table stay fixed at $-1$ as $x \to -1^-$. Therefore $\displaystyle\lim_{x \to -1^-} f(x) = -1$.

From part (a), $\displaystyle\lim_{x \to -1^+} f(x) = 1$, and from part (b), $\displaystyle\lim_{x \to -1^-} f(x) = -1$.

Since the left-hand limit and right-hand limit are not equal ($-1 \neq 1$), the two-sided limit $\displaystyle\lim_{x \to -1} f(x)$ does not exist.

The right-hand limit found in part (a) was one, which is positive. This occurred because for x slightly greater than negative one, the function simplified to the constant value one. Since this value stays fixed and never changes sign as x approaches negative one from the right, the answer is positive.