Plugging in the terminal value, $x=4$, yields the indeterminate form $\frac 00$. To find this limit analytically, we will factor the numerator and denominator and simplify the fraction. In the numerator we have a difference of squares, and such a form always factors in the following way: $$a^2 - b^2 = (a+b)(a-b).$$ Applying this general formula to our example, we get $$x^2 - 16 = (x+4)(x-4).$$ Next, in the denominator, we can factor out a common factor of $x$ from each of the terms: $$x^2 - 4x = x(x-4).$$ With these factorizations, we can simplify the fraction and find the limit:

\begin{align*} \lim _{x \to 4} \frac{x^2 - 16}{x^2 - 4x} &= \lim _{x \to 4} \frac{(x+4)(x-4)}{x(x-4)} \enspace & \text{(factoring)} \\[.4 em] &= \lim _{x \to 4} \frac{x+4}{x} & \text{(canceling)} \\[.4 em] &= \frac 84 & \\[.4 em] &= 2. & \end{align*}

Hence, $$\lim _{x \to 4} \frac {x^2 - 16}{x^2 - 4x} = 2.$$ This example uses the factor and cancel method.

\begin{align*} \lim _{x \to 2} \frac{x^2 + 3x - 10}{x^3 - 8} &= \lim _{x \to 2}\frac{(x-2)(x+5)}{(x-2)(x^2 + 2x + 4)} \enspace & \text{(factoring)} \\[.4 em] &= \lim _{x \to 2} \frac{x+5}{x^2 + 2x + 4} & \text{(canceling)} \\[.4 em] &= \frac{2+5}{2^2 + (2\cdot 2) + 4} & \text{(plugging in)} \\[.4 em] &= \frac{7}{12}. \end{align*}

Hence, $$\lim _{x \to 2} \frac {x^2 + 3x - 10}{x^3 - 8} = \frac {7}{12}.$$

With these factorizations, we can simplify the fraction and find the limit:

\begin{align*} \lim _{x \to 4} \frac{4x^2 - 19x + 12}{6x^2 -19x -20} &= \lim _{x \to 4}\frac{(x-4)(4x-3)}{(x-4)(6x + 5)} \enspace & \text{(factoring)} \\[.4 em] &= \lim _{x \to 4} \frac{4x-3}{6x + 5} & \text{(canceling)} \\[.4 em] &= \frac{16-3}{24+5} & \text{(plugging in)}\\[.4 em] &= \frac{13}{29}. \end{align*}

Hence, $$\lim _{x \to 4} \frac {4x^2 -19x +12}{6x^2 -19x -20} = \frac {13}{29}.$$

To solve this limit problem, we will use the conjugate radical of the numerator, which is $\sqrt {x} + 2$. In order to maintain the value of the expression given in the problem, we multiply by one in the form of the conjugate radical over itself:

\begin{align*} \lim _{x \to 4} \frac{\sqrt{x}- 2}{x-4} &= \lim _{x \to 4} \frac{\sqrt{x} -2}{x-4}\cdot \frac{\sqrt{x} +2}{\sqrt{x}+2} \\[.4 em] &= \lim _{x \to 4} \frac{x-4}{(x-4)(\sqrt{x}+2)} \\[.4 em] &= \lim _{x \to 4}\frac{1}{\sqrt{x}+2} \\[.4 em] &= \frac 14. \end{align*}

In the last step, we plugged in the terminal value $x=4$ to get the final answer $\frac 14$.

To solve this limit problem, we will use the conjugate radical of the numerator, which is $\sqrt {9+h}+3$. In order to not change the value of the expression given in the problem, we multiply by one in the form of the conjugate radical divided by itself:

\begin{align*} \lim _{h \to 0} \frac{\sqrt{9+h}-3}{h} &= \lim _{h \to 0} \frac{\sqrt{9+h}-3}{h}\cdot \frac{\sqrt{9+h}+3}{\sqrt{9+h}+3} \\[.4 em] &= \lim _{h \to 0} \frac{(9+h) - 9}{h(\sqrt{9+h} + 3)} \\[.4 em] &= \lim _{h \to 0} \frac{h}{h(\sqrt{9+h} + 3)} \\[.4 em] &= \lim _{h \to 0}\frac{1}{\sqrt{9+h} + 3} \\[.4 em] &= \frac 16. \end{align*}

In the last step, we plugged in the terminal value $h=0$ to get the final answer $\frac 16$.

\begin{align*} \lim _{x \to -3} \frac{x^2 - 9}{4 - \sqrt{13 -x}} &= \lim _{x \to -3} \left (\frac{x^2 - 9}{4 - \sqrt{13 -x}}\right ) \cdot \left (\frac{4 + \sqrt{13 -x}}{4 + \sqrt{13 -x}}\right ) \\[.4 em] &=\lim _{x \to -3} \frac{(x^2 - 9)\left (4 + \sqrt{13 -x}\right )}{x+3}\\[.4 em] &=\lim _{x \to -3} \frac{(x+3)(x-3)\left (4 + \sqrt{13 -x}\right )}{x+3} \\[.4 em] &= \lim _{x \to -3} (x-3)\left (4 + \sqrt{13 -x}\right )\\[.4 em] &= (-3-3)\left (4 + \sqrt{13- (-3)}\right ) \\[.4 em] &= (-6)(4 + 4) \\[.4 em] &= -48. \end{align*}

\begin{align*} \lim _{x \to 4} \frac{\frac{1}{x} - \frac{1}{4}}{x-4} &= \lim _{x \to 4} \frac{\frac{4-x}{4x}}{x-4}\\[.4 em] &=\lim _{x \to 4} \frac{4-x}{4x(x-4)}\\[.4 em] &= \lim _{x \to 4} -\frac{1}{4x} \\[.4 em] &= -\frac{1}{16}. \end{align*}

It is important to note that $a-b$ and $b-a$ are opposites which cancel, leaving $-1$: $$\frac {a-b}{b-a} = -1.$$ The ratio of opposites is $-1$.

Now back to the original problem: \begin{align*} \lim _{x \to -2} \frac{\frac{2}{x-3} + \frac{x+4}{5}}{x^2 + 5x + 6} &= \lim _{x \to -2} \frac{\frac{(x-1)(x+2)}{5(x-3)}}{(x+2)(x+3)}\\[.4 em] &= \lim _{x \to -2} \frac{(x-1)(x+2)}{5(x-3)(x+2)(x+3)} \\[.4 em] &= \lim _{x \to -2} \frac{(x-1)}{5(x-3)(x+3)}\\[.4 em] &= \frac{-3}{5(-5)(1)} = \frac{3}{25}. \end{align*}

Since $x \to 3^-$, we have $x<3$ and hence $x-3 <0$. Since the quantity in the absolute value bars is negative, we can compute its absolute value as follows:

$$|x-3| = -(x-3).$$

Using this in the limit, we get

\begin{align*} \lim _{x \to 3^-} \frac{|x-3|}{x-3} &= \lim _{x \to 3^-} \frac{-(x-3)}{x-3} \\[.4 em] &= \lim _{x \to 3^-} (-1) \\ &= -1. \end{align*}