First, multiply the numerator and denominator by $\displaystyle x-\sqrt{x+6}$ and it turns into

$\displaystyle \lim_{x \to -2} \frac{(x+\sqrt{x+6})}{(x+2)}\frac{(x-\sqrt{x+6})}{(x-\sqrt{x+6})}$

$\displaystyle \lim_{x \to -2} \frac{x^2-x-6}{(x+2)(x-\sqrt{x+6})}$

Factor the numerator and then cancel out the 'x+2'

$\displaystyle \lim_{x \to -2} \frac{(x+2)(x-3)}{(x+2)(x-\sqrt{x+6})}$

$\displaystyle \lim_{x \to -2} \frac{(x-3)}{(x-\sqrt{x+6})}$

After taking the limit, the answer is $\displaystyle \frac{5}{4}$

Factor the numerator

$\displaystyle \lim_{x \to 3} \frac{(x^2-\sqrt{a})(x^2+\sqrt{a})}{x^2-\sqrt{a}}=18$

Cancel the $\displaystyle x^2-\sqrt{a}$, plug in the limit and then solve for 'a'

$\displaystyle \lim_{x \to 3} x^2+\sqrt{a}=18$

$\displaystyle 3^2+\sqrt{a}=18$

$\displaystyle a=81$

We have the origin $\displaystyle (0,0)$ and a point $\displaystyle (x,y)$ located on the line.  That point represents the minimum distance to the orgin.  Apply the distance formula to these two points,

$\displaystyle d=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$

Plug in the line equation, take the derivative, set it equal to zero, and solve for x.

$\displaystyle d=\sqrt{x^2+(-\frac{1}{2}x+2)^2}=\sqrt{\frac{5}{4}x^2-2x+4}$$\displaystyle {d}'=\frac{\frac{5}{2}x-2}{2\sqrt{\frac{5}{4}x^2-2x+4}}=0$

$\displaystyle x=\frac{4}{5}$

Use this $\displaystyle x$ value to find $\displaystyle y$

$\displaystyle y=-\frac{1}{2}\left ( \frac{4}{5} \right )+2=\frac{8}{5}$

So we have the point $\displaystyle (\frac{4}{5},\frac{8}{5})$, which is closest to the origin. We can now find its distance from that origin.

$\displaystyle d=\sqrt{\left (\frac{4}{5} \right )^2+\left ( \frac{8}{5} \right )^2}$

$\displaystyle d=\sqrt{\frac{16}{5}}=\frac{4}{\sqrt{5}}$

First, factor the numerator of the integrand.

$\displaystyle \int_{-5}^{5}\frac{(x+6)(x+c)}{x+6}dx$

Cancel out $\displaystyle x+6$

$\displaystyle \int_{-5}^{5}\frac{(x+6)(x+c)}{x+6}dx=-60$

$\displaystyle \int_{-5}^{5}(x+c)dx=-60$

Perform the integral and then solve for $\displaystyle c$

$\displaystyle \frac{1}{2}(5)^2+5c-\left ( \frac{1}{2}(-5)^2+(-5)c \right )=-60$

$\displaystyle 10c=-60$

$\displaystyle c=-6$

Taking the derivative of an integral yields the original function, but because we have a different variable in the integration limits, the variable switches

using integration identities: $\displaystyle \int{x^n dx}= \frac{x^{n+1}}{n+1} and \int{sin(x) dx}= - cos(x) and \int{csc^2(x) dx}= -cot (x)$ 

$\displaystyle \int_{1}^{4}(4x^2+2x-8)dx$

$\displaystyle \frac{4x^3}{3} + \frac{2x^2}{2} -8x$

$\displaystyle \frac{4x^3}{3} + x^2 -8x$         evaluate at $\displaystyle \left [ 1,4 \right ]$

$\displaystyle \left(\frac{4(4)^3}{3} + (4)^2 -8(4)\right) - \left(\frac{4(1)^3}{3} + (1)^2 -8(1)\right)$

$\displaystyle =75$

$\displaystyle \int_{0}^{3}(2x(x^2+1)^3)dx$

Intergation by substitution

$\displaystyle u= x^2+1$

$\displaystyle du=2xdx$

new endpoints:

$\displaystyle u=(3)^2+1=10$

$\displaystyle u=(0)^2+1=1$

New Equation:

$\displaystyle \int_{1}^{10}(u^3)du$

$\displaystyle \frac{u^4}{4}$ at $\displaystyle \left [ 1,10 \right ]$

$\displaystyle = \frac{(10)^4}{4} - \frac{(1)^4}{4}$

$\displaystyle = \frac{(10000)}{4} - \frac{(1)}{4}$

$\displaystyle = \frac{9999}{4}$

The relationship between $\displaystyle ln(x)$ and x is an exponential relationship; $\displaystyle x$ is going to $\displaystyle 0$ exponentially faster than $\displaystyle ln(x)$ is going to $\displaystyle -\infty$ . One way to prove this is to write $\displaystyle x = 1/(1/x)$ and use L'Hôpital's rule:

$\displaystyle \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{\ln'x}{(1/x)'} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{-x^2}{x} = -0 = 0$

The rational function $\displaystyle \frac{4x^3 - 21 x + 10}{x^2 - 5x + 6}$  has a denominator with two roots, $\displaystyle x=2$ and $\displaystyle x=3$. These are discontinuities.

Factoring both top and bottom and canceling a term $\displaystyle x-2$ tells us that this function is equal to 

$\displaystyle \frac{(2x - 1) (2x + 5)}{x - 3}$ 

except at $\displaystyle x=2$. This point is a removable discontinuity. $\displaystyle x=3$ is therefore an essential discontinuity where the ratio goes to $\displaystyle \pm \infty$.