Answer:
d) The limit does not exist
General Formulas and Concepts:
<u>Calculus</u>
Limits
- Right-Side Limit:

- Left-Side Limit:

Limit Rule [Variable Direct Substitution]: 
Limit Property [Addition/Subtraction]: ![\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20c%7D%20%5Bf%28x%29%20%5Cpm%20g%28x%29%5D%20%3D%20%20%5Clim_%7Bx%20%5Cto%20c%7D%20f%28x%29%20%5Cpm%20%5Clim_%7Bx%20%5Cto%20c%7D%20g%28x%29)
Step-by-step explanation:
*Note:
In order for a limit to exist, the right-side and left-side limits must equal each other.
<u>Step 1: Define</u>
<em>Identify</em>

<u>Step 2: Find Right-Side Limit</u>
- Substitute in function [Limit]:

- Evaluate limit [Limit Rule - Variable Direct Substitution]:

<u>Step 3: Find Left-Side Limit</u>
- Substitute in function [Limit]:

- Evaluate limit [Limit Rule - Variable Direct Substitution]:

∴ Since
, then 
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits
Answer:
5 withdrawals
Step-by-step explanation:
Since $10.50 ends in $0.50, you know that she must have made at least one $2.50 transaction.
10.50 - 2.50 = 8
8/2 = 4 (she made four $2 transactions)
She made 4 $2 transactions and 1 $2.50 transaction.
4 + 1 = 5
She made 5 transactions total.
Part A :
80x0.18= 14.4
Part B :
80x 0.075= 6
Part C
14.4+6 +80=$100
Assuming that there are 8 names that are going to be used for a random order;
First, let's try to figure out how many possible ways the names can be arrange:
=8 choices x 7 choice x 6 choices x 5 ... x 1 choice (if you select person A to go first, they cannot be second as well, so that is why 8 choices is multiplied by 7 instead of 8)
This can also be written as 8!
The specific way that a random set of names is chosen by you and actually chosen is 1. Each name has a specific order, in each term so there can only be one possible way.
Therefore, P(order being chosen)=1/8!
Hope I helped :)