welcome back everyone today we will be going over unit 1 of calculus a b which is on limits and continuity there are definitely a few questions from unit 1 on the AP exam for both mcqs and frqs make sure to subscribe to know when I release unit 2 and also visit my channel to find problem walkthroughs let's start by defining what is a limit and what does it mean here we have a graph of y equals 3x we can say that the limit of f of x as X approaches 1 is 3. this is because as the X values approach one the Y values converge towards three now if we have a piecewise function where a point doesn't exist at 1 3 the limit is still three this is because the graph continues to approach three when X approaches one limits are all about the point being approached limits can also be one-sided and these are notated by A Plus or a minus sign next to the value that X approaches a minus sign means from the left and the plus sign means from the right in this scenario the limit from the left is four and limit from the right is six since the limit from both sides are different the limit as X approaches one does not exist sometimes you'll be asked to add or multiply limits and as you can see here they're really intuitive for example you could be given two graphs and we find the limit for each of the functions and then we just need to add them to solve the most common type of limit problems will be solving them analytically with simple limits such as x squared minus three you can just plug in x equals 2 and find our answer one but what happens if there is a division over zero if we get a number that is not zero over zero we can plug in the numbers to find the limit on both sides if it's a positive number the limit is infinity and otherwise it's negative Infinity if the two numbers match you have your answer but if they don't then the limit does not exist in this example of zero over zero we can simplify the limit by canceling out the factor and get our answer five in other examples canceling out may not be so simple here we have to multiply by a conjugate to manipulate the numbers we can do this because it is basically multiplying by one over one after multiplying they cancel out and we can plug in the value and find the answer for trig limits we use the same steps as above but we use trig identities when possible here are the most common ones there are also special rules for trig limits to zero specifically in this example sin 5x over X we would multiply by five over five we can apply the rule to cancel out the sin 5x over 5x part and get our answer 5. similarly if you ever spot one minus cosine X over X the answer would just be zero The Squeeze theorem is also very useful for solving limits especially trigonometric ones it states if f of x is less than or equal to G of X is less than or equal to H of X the same also applies to their limits when given the sign limit we know that the value of sine is only bounded by negative one to one so we can set them in an equation we can multiply all three parts by the outside value as well and at the end we just take a limit of all three we end up with the answer of zero when a limit equals infinity this means that there is a vertical asymptote the function continues increasing forever and never reaches the point this is shown in the graph however limits to Infinity are horizontal asymptotes in this instance the graph approaches 4 as X approaches Infinity but will never reach it we can solve limits to Infinity using three simple rules the first step is identifying the growth rate of the top and the bottom for example an exponential function grows faster than a polynomial function it's a growth rate on the bottom is larger the answer is zero examples of this include x squared over X cubed or x to the power of four over e to the power of x secondly if the growth rates are the same we take the fraction of the leading coefficients the last rule covers if a larger growth rate is on top we plug in a large positive number if it's positive infinity and the large negative number for negative Infinity the sign of the answer dictates if the answer is positive or negative Infinity the other topic of this unit is continuity when something is continuous it does not break here are three types of discontinuities the first is at a point where a function has a limit but there's no point at the value of the limit it can be somewhere else or it doesn't exist at all in jumps the limit from both sides are different and there is a large space in between them and it's kind of like a jump in asymptotes the limits go to infinity and are never able to reach each other continuity is guaranteed when the limit equals the point given this piecewise function we can test its continuity by finding F of 2 we plug in x equals 2 to the function that has two in its domain we find the limit from the left and from the right and they are different the limit does not exist and this does not equal to that means the piecewise function is not continuous another common question is finding a value to make a piecewise function continuous in this example we know that continuity requires limits from both sides to be equal we set them equal to each other plug our values in and our answer is k equals negative one the ivt which is also known as the intermediate value theorem is a theorem that involves continuity it says that if a function f of x is continuous on the interval A and B that includes every y value between F of a and F of B at least once in this example problem all three intervals contain negative one because it is between 1 and negative two negative two and three and three and negative five so that was it for unit one limits and continuity on the AP exam are pretty simple so you should just make sure to remember the basic rules I have gone over if there's anything else I missed where you don't fully understand feel free to leave a comment and I can help explain it to you thanks for watching And subscribe to know when I post a video on unit 2.