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 \(\displaystyle \pm \infty\).