$$ \lim_{x \to 1} \frac{x^3 - 1}{x - 1} $$

$\to$ = "approaches"

Factoring the expression, we get $x^2 + x + 1$

This expression returns 3 when $x = 1$, while it does not exist in the original function. However, limits do not care whether or not the value actually exists, but rather where it would be approaching. Therefore,

$$ \lim_{x \to 1} \frac{x^3 - 1}{x - 1} = 3 $$

$$ \lim_{x \to 0} \frac{x}{\sqrt{x + 1} - 1} $$

When $x = 0$, we get $\frac{0}{0}$, which is not undefined, but indeterminate. This means we have to find other methods for solving this problem.

$$ \lim_{x \to 0} \frac{x}{\sqrt{x + 1} - 1} \cdot \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1} = \lim_{x \to 0} \frac{x(\sqrt{x + 1} + 1)}{x + 1 - 1} = \lim_{x \to 0} (\sqrt{x + 1} + 1) = 2 $$

$$ \lim_{x \to 0} f(x) $$

The graph gradually increases from $[-2, 0)$ and decreases from $(0, 2]$. That means there is an asymptote at 0; therefore,

$$ \lim_{x \to 0} f(x) = \textrm{DNE} $$

$f(x) = $

$$\left\{ \begin{array}{ll} 1 & x \neq 2 \\ -1 & x = 2 \end{array} $$

$$ \lim_{x \to 2} f(x) $$

When graphed, it would be the line $y = 1$ with a removable discontinuity at $x = 2$.

\[ \lim_{x \to 0} \frac{x}{|x|} \]

The left hand limit is -1, but the right hand limit is 1. Therefore,

\[ \lim_{x \to 0} \frac{x}{|x|} = \textrm{DNE} \]

\[ \lim_{x \to 0} \frac{1}{x^2} \]

\[ \lim_{x \to 0^{+}} \frac{1}{x^2} \] The limit from the left approaches $\infty$. The limit from the right approaches $\infty$. Therefore,

\[ \lim_{x \to 0} \frac{1}{x^2} = \infty \]

\[ \lim_{x \to 0} sin(\frac{1}{x}) \]

When graphed, the frequency of the oscillations increase as you approach 0, and it is impossible to determine the limit from the graph.