Prove $\displaystyle{\lim_{x \rightarrow -1} (5x+7) = 2}$ using the epsilon-delta definition of a limit.
Choose $\delta = \epsilon/5$.
Prove $\displaystyle{\lim_{x \rightarrow 5} (x^2 - 3x) = 10}$ using the epsilon-delta definition of a limit.
Choose $\delta = \textrm{min}\left\{3,\epsilon / 10\right\}$ (solution with annotated work)
Prove $\displaystyle{\lim_{x \rightarrow 2} (x^2 - 5x + 6) = 0}$ using the epsilon-delta definition of a limit.
Choose $\delta = \textrm{min}\left\{1,\epsilon / 2\right\}$
Sketch a graph of a function satisfying the following conditions, then define a piecewise-defined function that satisfies the same:
$$\lim_{x \rightarrow -2} f(x) = 1; \quad \lim_{x \rightarrow 1^-} f(x) = -2; \quad \lim_{x \rightarrow 1^+} f(x) = 4; \quad f(-2) = 0; \quad f(1) = -4$$Answers vary.
Determine the infinite limit: $\displaystyle{\lim_{x \rightarrow 4^-} \frac{\sqrt{x} - 1}{(x^2 - 16)^5}}$
As $x \rightarrow 4^-$, the numerator has a limiting value of 1, while the denominator can be made arbitrarily small and negative. Thus, their quotient is negative and can be made arbitrarily large in magnitude. Thus, there is a vertical asymptote at $x = 4$ (on the left), and this limit fails to exist.