Unit 1 Math Review

Note: not saying these show all the work we’re supposed to show, honestly idk what we’re supposed to show lmao. They’re just like, explaining the concept I guess. Also, the answer is just provided straight up at the top so you don’t have to scroll through it all to get to it if that’s really all you want.

Some questions are not in order (I grouped them by topic instead), and not all of them are here, but I think all the ones that touch on the concepts the most are.

Limit Evaluation §

Limit stuff with infinity in it §

3. Determine the infinite limit. §

$$\underset{x\to 2^{+}}{\lim }\frac{x+9}{x-2}=\infty$$

What we know:

1. Subbing in 2 gives us an indeterminate $\frac{11}{0}$ value.
2. A non-zero value over 0 yields an asymptote.
3. We are approaching from the right or positive side; therefore, $x>2$.

If $x>2$, then $(x-2)>0$, and $(x+9)>11$. As the numerator gets infinitely bigger and the denominator gets infinitely smaller, the value just keeps getting bigger as it approaches 2.

$$\frac{\text{bigger}}{\text{smaller}}=\text{keeps getting bigger}$$

Note

This would also be true if the denominator was infinitely getting bigger, as long as it got bigger slower than the numerator did.

4. Determine the infinite limit. §

$$\underset{x\to 9^{-}}{\lim }\frac{x^2-9x}{x^2-18x+81}=-\infty$$

What we know:

1. Subbing in 9 gives us an indeterminate $\frac{0}{0}$ value.

Before we continue listing what we know, it is beneficial to cancel this out a bit so we can get a better look at it. It is pretty easy to simplify:

$$\begin{array}{r}\underset{x\to 9^{-}}{\lim }\frac{x^2-9x}{x^2-18x+81}\\ \\ \underset{x\to 9^{-}}{\lim }\frac{x(x-9)}{(x-9)(x-9)}\\ \\ \underset{x\to 9^{-}}{\lim }\frac{x}{x-9}\end{array}$$

While substituting 9 in this new equation only leads us to another indeterminate, $\frac{9}{0}$, this time we see that it is a non-zero over zero, which indicates an asymptote like we saw before.

Now, what we know:

1. Subbing in 9 gives us an indeterminate $\frac{9}{0}$ value.
2. A non-zero value over 0 yields an asymptote.
3. We are approaching from the left or negative side; therefore, $x<9$.

If $x<9$, then $(x-9)<0$. AKA, $x-9$ is NEGATIVE.

$x-9$ is obviously less than x (9 less than x…), so we know that $\frac{\text{more}}{\text{less}}$ keeps getting bigger— infinitely bigger as we approach 9 from the left side. Infinite and negative gives us $-\infty$.

14. Evaluate the limit using the appropriate properties of limits. §

$$\underset{x\to \infty }{\lim }\frac{8x^2-9}{3x^2+x-3}=\frac{8}{3}$$

Because we’re approaching infinity, we know that $x^2$ grows so much faster than $x$ or any constant that everything else basically becomes irrelevant as we’re talking on such a big scale. So, what we’re really looking at is:

$$\underset{x\to \infty }{\lim }\frac{8x^2}{3x^2}=\frac{8}{3}$$

15. Find the limit, if it exists. §

$$\underset{x\to -\infty }{\lim }\frac{x-9}{x^2+2}=0$$

Again, since we’re approaching an infinity, it is a matter of rate of growth. $x^2$ grows faster than $x$, and the numbers quickly become so big that the constants and plain $x$ become irrelevant compared to the growth of $x^2$, so what we’re really looking at when we put on the large scale gloves is:

$$\underset{x\to -\infty }{\lim }\frac{1}{x^2}$$

When you square something, the negative goes poof, and we know that $x^2$ keeps getting bigger and bigger; $\frac{\text{smaller}}{\text{bigger}}$ keeps getting INFINITESIMALLY smaller, aka… approaching 0!

16. Find the limit, if it exists. §

$$\underset{t\to \infty }{\lim }\frac{\sqrt{t}+t^2}{9t-t^2}=-1$$

Like before, we know that $t^2$ gets so much bigger so much quicker than $\sqrt{t}$ or $9t$, making them irrelevant for our consideration. So, in reality, what we’re seeing is a much easier equation:

$$\underset{t\to \infty }{\lim }\frac{t^2}{-t^2}=-1$$

17. Find the limit, if it exists §

$$\underset{x\to \infty }{\lim }\frac{x^5}{\sqrt{x^{10}+2}}=1$$

Again… we’re approaching infinity, so the $+2$ inside $\sqrt{x^{10}+2}$ becomes irrelevant as our scale increases and increases. So, we’re now looking at:

$$\underset{x\to \infty }{\lim }\frac{x^5}{\sqrt{x^{10}}}$$

If you recall square what a square root actually is, you remember that finding the square root of something is essentially raising it to the $\frac{1}{2}$ power:

$$\sqrt{x^{10}}=x^{{10}^{\frac{1}{2}}}=x^{10\times \frac{1}{2}}=x^5$$

Soooooo…

$$\underset{x\to \infty }{\lim }\frac{x^5}{x^5}=1$$

19. Find the limit, if it exists. §

$$\underset{x\to \infty }{\lim }(e^{-8x}\cos (x))=0$$

Ew, pre-calc flashbacks. I never want to touch a cosine function ever again in my life. Good news: $e^{-8x}$ grows (well, more like shrinks) so much faster than $\cos (x)$ that it is ridiculous, so we can just throw $\cos (x)$ out and instead look at this beautiful, trig-less limit instead:

$$\underset{x\to \infty }{\lim }\frac{1}{e^{8x}}$$

So, it’s getting infinitesimally smaller… yeah, by this point we know the drill.

Stuff… without infinity in it §

6. Evaluate the limit, if it exists. §

$$\underset{x\to 3}{\lim }\frac{x^2-5x+6}{x-3}=1$$

This just makes you use basic simplification.

$$\begin{array}{r}\underset{x\to 3}{\lim }\frac{(x-2)(x-3)}{(x-3)}\\ \\ \underset{x\to 3}{\lim }(x-2)\end{array}$$

Then you just sub in 3, to get $3-2=1$.

7. Evaluate the limit, if it exists. §

$$\underset{h\to 0}{\lim }\frac{\sqrt{81+h}-9}{h}=\frac{1}{18}$$

For this one, you just have to use the conjugate to remove the square root, so that you can simplify out the h.

$$\begin{array}{rl}\underset{h\to 0}{\lim }\frac{\sqrt{81+h}-9}{h}& =\underset{h\to 0}{\lim }\frac{\sqrt{81+h}-9}{h}\times \frac{\sqrt{81+h}+9}{\sqrt{81+h}+9}\\ & =\underset{h\to 0}{\lim }\frac{(81+h)-81}{h(\sqrt{81+h}+9)}\\ & =\underset{h\to 0}{\lim }\frac{h}{h(\sqrt{81+h}+9)}\\ & =\underset{h\to 0}{\lim }\frac{1}{\sqrt{81+h}+9}\end{array}$$

Then, when you substitute in the zero:

$$\underset{h\to 0}{\lim }\frac{1}{\sqrt{81+h}+9}=\frac{1}{9+9}=\frac{1}{18}$$

8. Find the limit, if it exists. §

$$\underset{x\to 1}{\lim }\frac{10x+10}{\mid x+1\mid }=\text{DNE}$$

Since there is an absolute value here, we need to evaluate the limit from both sides.

$$\begin{array}{rlrl}& \text{From the left:}& & \text{From the right:}\\ & \underset{x\to -1^{-}}{\lim }\frac{10x+10}{-(x+1)}& & \underset{x\to -1^{+}}{\lim }\frac{10x+10}{x+1}\\ & =\underset{x\to -1^{-}}{\lim }\frac{10(x+1)}{-(x+1)}& & =\underset{x\to -1^{+}}{\lim }\frac{10(x+1)}{x+1}\\ & =\underset{x\to -1^{-}}{\lim }-10& & =\underset{x\to -1^{+}}{\lim }10\\ & =-10& & =10\end{array}$$

The left and right side limits do not equal each other as $-10\ne 10$, so the limit does not exist.

Discontinuities §

continuity has some stuff about this too, but it is kind of cringe.

10. Explain why the function is discontinuous at the given number a. §

$$f(x)=\frac{1}{x+5},a=-5$$

Let’s test this with our continuity conditions. Note that this question wants all that apply, so we need to go through all of them, as opposed to stopping at the first failure like we’d normally do.

Condition $i$; does ${\lim }_{x\to a}f(x)$ exist? §

Let’s consider both sides of ${\lim }_{x\to -5}\frac{1}{x+5}$, because obviously subbing in -5 just brings us out to an indeterminate $\frac{1}{0}$. We do know that $\frac{1}{0}$ is a non-zero/zero which means we’re approaching an asymptote, but we want to know if we are going to the same infinity from both sides.

For ${\lim }_{x\to -5^{-}}\frac{1}{x+5}$, we are approaching from the left or negative side; therefore, $x<-5$. So, if x is always smaller than -5, $(x+5)<0$ always as well. This means that our denominator is negative, and that from the left we approach $-\infty$ for this asymptote.

For ${\lim }_{x\to -5^{+}}\frac{1}{x+5}$, we apply the same logic in reverse. $x>-5$, so $(x+5)>0$. Therefore, the denominator is positive, and we approach $\infty$ at $x=-5$.

This also means that ${\lim }_{x\to -5}f(x)$ does not exist, so it fails condition $i$, but since we need to check off all conditions we of course have to keep going…

Condition $ii$; does $f(a)$ exist? §

This is my favorite condition, because it is really easy to do.

$$\begin{array}{r}f(x)=\frac{1}{x+5}\\ \\ f(-5)=\frac{1}{-5+5}\\ \\ \\ f(-5)=\frac{1}{0}\end{array}$$

Nope. Does not exist. Can’t divide by zero, so it fails condition $ii$ as well.

Condition $iii$; does ${\lim }_{x\to a}f(x)=f(a)$? §

I mean… I think so? Yes, I guess? They both don’t exist…? But this isn’t even an option on the WebAssign so it doesn’t really matter for this question, I don’t think we have to know this.

11. Explain why the function is discontinuous at the given number a. §

$$f(x)=\{\begin{array}{l}\frac{x^2-5x}{x^2-25}\text{if}x\ne 5\\ \\ 1\text{ if }x=5\end{array}$$

Now we have to test the conditions again:

Condition $i$; does ${\lim }_{x\to a}f(x)$ exist? §

Yes; from both sides, until $x=5$, the limit exists. It’s easier to see if you simplify:

$$\begin{array}{r}\underset{x\to 5}{\lim }\frac{x(x-5)}{(x+5)(x-5)}\\ \\ \underset{x\to 5}{\lim }\frac{x}{x+5}\\ \\ \frac{5}{5+5}\\ \\ \underset{x\to 5}{\lim }\frac{x^2-5x}{x^2-25}=\frac{1}{2}\end{array}$$

Condition $ii$; does $f(a)$ exist? §

Yes; $f(5)=1$, as defined in the piecewise function above.

Condition $iii$; does ${\lim }_{x\to a}f(x)=f(a)$? §

No, because $\frac{1}{2}\ne 1$.

Derivatives and Friends §

20. Consider the parabola $y=8x-x^2$ §

a. Find the slope of the tangent line to the parabola at point $(1,7)$. §

The slope of the tangent line is the derivative, so we have to go through the whole shebang…

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{[8(x+h)-(x+h{)}^2]-(8x-x^2)}{h}\\ & =\underset{h\to 0}{\lim }\frac{8x+8h-x^2-2xh-h^2-8x+x^2}{h}\\ & =\underset{h\to 0}{\lim }\frac{8h-2xh-h^2}{h}\\ & =\underset{h\to 0}{\lim }(8-2x-h)\\ f^{\prime }(x)& =8-2x\end{array}$$

So, substituting in our $x$ value of 1:

$$\begin{array}{rl}{f}^{\prime }(1)& =8-2(1)\\ f^{\prime }(1)& =6\end{array}$$

The slope of the tangent line at $(1,7)$ is 6.

b. Find the equation of the tangent line in part a. §

Then, to get the equation of the tangent line, we substitute our slope and point into point-slope form:

$$\begin{array}{r}\text{Point-slope form: }y-y_1=m(x-x_1)\\ \text{Substituting: }y-7=6(x-1)\\ y-7=6x-6\\ y=6x+1\end{array}$$

The tangent line at point $(1,7)$ is $y=6x+1$. WebAssign hates us and wants the $y=$ as well, so don’t forget that when inputting it.

21. Find the equation of the tangent line at the given point. §

$$y=\sqrt{x},(81,9)$$

First we’re just going to find the slope of the tangent line at $(81,9)$— aka, the derivative with the point subbed in.

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}\\ & =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\ & =\underset{h\to 0}{\lim }\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\\ & =\underset{h\to 0}{\lim }\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\\ & =\underset{h\to 0}{\lim }\frac{1}{\sqrt{x+h}+\sqrt{x}}\\ & =\frac{1}{2\sqrt{x}}\end{array}$$

So, now that we’ve found the derivative, we can just substitute our $x$ value of 81 in to get the slope at $x=81$.

$$\begin{array}{rl}{f}^{\prime }(81)& =\frac{1}{2\sqrt{81}}\\ & =\frac{1}{2(9)}\\ & =\frac{1}{18}\end{array}$$

Now that we know the slope, we can put it into the good old fashioned point-slope function like so:

$$\begin{array}{r}\text{Point-slope form: }y-y_1=m(x-x_1)\\ \text{Substituting: }y-9=\frac{(x-81)}{18}\\ 18(y-9)=x-81\\ 18y-162=x-81\\ 18y=x-81+162\\ 18y=x+81\\ y=\frac{x+81}{18}\end{array}$$

Anddd we’re done. Finally.