in section 2.4 the topic is infinite limits so in an infinite limit the dependent variable in other words the y value becomes boundless either positive or negative as the independent variable or the x value approaches some finite value and so by boundless what we mean is either positive infinity or negative infinity so the output of the function is growing extremely large either in the positive direction or the negative direction so for example if as x goes to c f of x goes to infinity we're going to write the following statement the limit as x goes to c of f of x is equal to infinity so back in section 2.1 or 2.2 maybe it was we talked about some cases where limits failed to exist and one of those cases was when the function's output grows too large and never approaches a particular value so technically such a limit as this does not exist however we're interested in functions that behave this way because it gives us some insight into the behavior of the function and so we are going to study limits of this form and we are going to sort of bend our rules a little bit and in this case we're going to write limits equal to the value of infinity or negative infinity so again we're going to just neglect that fact because we're interested in limits that behave we're interested in functions that behave this way because it gives us some insight into the behavior of the function so when will an infinite limit occur an infinite limit is going to occur when we apply direct substitution and we have a non-zero in the numerator and a zero in the denominator so we're going to have infinite limits when we have a zero in the denominator only so let's take a a look here at a particular example so here we have the function f of x equals 1 over x and let's observe the behavior of the function's output as we approach zero from the right hand side so approaching zero from the right hand side let's select some inputs here so i have one tenth one one hundredth one one thousandth and so on so notice the trend here we're approaching zero from values larger than zero so we're seeing a right hand approach to zero as we approach zero on the right what's happening to our outputs you should very clearly see our outputs here are growing very large so our outputs go from one to ten to a hundred to a thousand to ten thousand to a hundred thousand all right so what we're seeing here if we continue plugging in smaller and smaller and smaller numbers approaching 0 this output value is going to continue to get larger and larger and larger so in limit terms what we're seeing here we're seeing this limit the limit as x goes to 0 on the right of 1 over x is equal to infinity and if we were to do the same thing on the left hand side of zero we'll have the limit as x goes to zero on the left of one over x equals negative infinity all right so notice that direct substitution of the zero gives me a one over a zero so a zero in the denominator only when that occurs we're going to have an infinite limit either plus or minus infinity so numerically we're done we're demonstrating this fact here as x approaches 0 on the right the y values are approaching infinity so here's a graph of that function y equals 1 over x or f of x equals 1 over x so you'll see as we go to 0 on the right our curve is going up to infinity as we go to zero on the left our curve is going down to negative infinity so with these limits we have some difficulty notationally there's no good and compact way of expressing our way through these types of limits so we make some conventions that we just accept uh they're not necessarily the only way this might be notated it's not necessarily purely correct mathematical notation it may be a little bit of an abusive notation but this is how we will work our way through limits of this form so we're going to let c be a real number and in this case we're just taking c to be positive so we're going to use the following notation for this style of limits so we will take c divided by a small positive to be positive infinity and c divided by a small negative to be negative infinity and so our normal sine rules will apply if i have a negative on the top negative over positive is of course negative if i have a positive here on the top positive over negative is of course negative and and um negative over negative is positive so all the normal sign rules will apply here but for the sake of simplicity i'm just going to express it uh with the assumption that c is larger than zero so again this is our notation for this type these types of limits uh we'll see how we apply it so again it may be a little bit strange because we're going to be writing you know this statement small positive small negative in the middle of some mathematical argumentation so we just accept this as the way we will approach these limits and sort of neglect the fact that we perhaps are abusing uh notation here so let's look back through our limit strategies let's see how we will approach limits uh you know up to this section 2.4 if direct substitution gives us a real number of course we're finished if direct substitution gives that zero over zero indeterminate form the limit will typically exist and so we need to work through some of those techniques that we discussed in section 2.3 and finally up to section 2.4 now if a if direct substitution gives a zero in the denominator only then we're going to have one of these infinite limits all right so again technically does not exist but we investigate these limits anyway so let's look at a couple of examples here i've got the limit as x goes to 2 on the left and the limit as x goes to 2 on the right of x plus 3 divided by x minus 2. so you will notice here that plugging in 2 i'm going to have a 5 over 0 in both instances so this is going to be one of our infinite limits so our goal is to find what exactly we have for each of these either plus infinity or minus infinity so i'm going to look at the first one the limit as x goes to 2 on the on the left so here's how we'll approach this we've got x plus 3 divided by x minus 2. so what we're going to do is we're going to go ahead and plug in 2 into the numerator of course that's going to be 5. and so for the bottom i've got to figure out on this approach to 2 the left-hand approach to 2 is the bottom approaching 0 through the negatives or through the positive so the way i typically do this is by finding a test point so if we take a test point let's say this is our value of 2 we're approaching 2 on the left hand side and so maybe a test point here is just 1. all right certainly there are more options you could pick 0 you could pick anything to the left of two so if i were to take this test point of one and plug it in to the bottom one minus two is of course negative one and so as we approach 2 on the left hand side we're approaching 0 right because 2 minus 2 is 0 but we're going to be approaching our test point shows us we're going to be approaching through the negatives so what i'm going to write here is 5 divided by small negative and our convention here 5 divided by small negative will be interpreted as negative infinity so likewise for the second limit the limit as x goes to 2 on the right of x plus 3 divided by x minus 2 of course plugging in 2 i'm going to have 5 again on the top the question now is small positive or small negative so now we're approaching 2 from the right so let's pick a test point let's say maybe 3 we're going to plug 3 into the bottom 3 minus 2 is one so we are positive and approaching zero so what i'm gonna write here is five divided by small positive and our convention is to interpret that as infinity so two very simple of these two very simple examples of these infinite limits again the most difficult thing is determining which direction i'm approaching zero from on the bottom and i'm approaching through the negatives in which case we'll write small negatives or am i approaching zero through the positives we'll write small positive and then again we remember our convention we interpret this to be positive or negative infinity applying the normal sine rules for division of numbers so here's another limit the limit as x goes to 0 on the right of x squared minus 3x plus 2x excuse me just plus 2 divided by x cubed minus 2x squared so direct substitution plug in zero we're going to have 2 on the top and 0 on the bottom so we have a 0 in the bottom only that means that this limit is going to be infinite so we want to figure out do we have plus infinity or do we have minus infinity so we've got the limit as x goes to 0 on the right of our function x squared minus 3x plus 2 divided by x cubed minus two x squared so we'll plug in zero into the numerator we're left with a two all right and so for the bottom we need to select a test point so we're approaching zero from the right so from values larger so an easy test point would just be one so plugging that one into the bottom i'm gonna have one cubed minus two times one squared well that's just going to be 1 minus 2 so 1 minus 2 is of course negative 1. so we're approaching 0 on the bottom through the negatives so we're going to say 2 divided by small negative and our convention is to interpret that as minus infinity all right and you're welcome to check these graphically if you graph this function on desmos or a graphing calculator you should note that as we approach zero from the right that this curve is going to be going down to negative infinity here we have the limit as x goes to zero on the right of one plus two x divided by x squared so again notice direct substitution is going to give me 1 divided by 0 so this limit is going to be infinite so we have to figure out is it plus or minus infinity so the limit as x goes to 0 on the right 1 plus 2x divided by x squared all right plugging in zero we'll have a one on the top now in this case things are fairly simple because i have a square in the bottom and i know a square is always positive so i don't have to really think much here the bottom will be small positive and of course our interpretation 1 divided by small positive will be infinity another limit i have the limit as j goes to 0 on the right of 8 to the j power divided by 1 minus 3 to the j all right so if i plug in 0 then on the top a 0 exponent is always one and on the bottom i'm going to have one minus one in that case so i'm gonna have one over zero so we're gonna have an infinite limit here so plugging in zero in the top we have our one and so again we just want to take a test point so to the right of zero an easy test point is one all right so into the bottom i'm going to have 1 minus 3 to the first power which is of course going to be 1 minus 3 which is going to be negative 2 so we're approaching 0 on the bottom through the negatives i have one divided by small negative which i interpret as negative infinity so this brings us to the topic of vertical asymptotes so we have a definition for vertical asymptotes here the line x equals c is called a vertical asymptote we may abbreviate v a of the function f if any one of the limits as x approaches c is infinite all right so this these infinite limits are giving us a way to characterize vertical asymptotes so for instance we say that the line x equals 0 is a vertical asymptote of this function f of x equals 1 over x since we've seen that the limit as x goes to zero on the right of one over x is infinity all right so we're going to be able to say now that vertical asymptotes will occur when the denominator only is equal to zero okay now regardless we will still want to verify that we have an infinite limit as we approach the particular x value that we believe there is a vertical asymptote existing at so however we will identify candidates for vertical asymptotes by looking for where the denominator only is equal to zero so i only have to show that one of these limits is infinite so either the right hand limit or the left-hand limit if any one of those limits is infinite then we can conclude that we have a vertical asymptote as we approach this particular input value or this particular x value all right so let's find and then verify the vertical asymptotes of this function f of x equals x plus one divide by two x squared plus x minus three so our first task is to find where the denominator is equal to zero so we will set the denominator two x squared plus x minus three equal to zero and so this is going to factor into two x plus three and x minus one so let's verify that by foiling here's 2x squared here's minus 2 x here's plus 3x that's positive x 3 times negative 1 is negative 3. so we've got two potential locations where the bottom will be zero so two x plus three equal to zero and then x minus one equal to zero so subtract three divide by two we've got x equals negative three halves add one we have x equals positive one so two potential vertical asymptotes at x equals negative three halves and x equals negative one excuse me positive one so what i'd like to do now is verify that these are going to be infinite as we approach those x values from some side so again these are just going to be at this stage potential vertical asymptotes and i want to verify through a limit that we have the value of plus or minus infinity on our approach so let's take the one approaching one first and so we can pick any direction the left hand direction or the right hand direction sometimes i get into the habit of just liking the right hand direction but there's nothing special about the right hand direction i could also evaluate this from the left hand direction so we'll plug in one into the numerator so that's of course going to give me a two on the bottom you see two times one is going to be two plus one is three take away three zero so we need to find a test point right so a test point let's just say we want to be to the right of one so a test point could just be 2 so we're approaching 1 from the right so we're going to plug in 2 here this is going to be 2 times 2 squared which is going to be 8 8 and 2 is 10 10 take away 3 is 7. anyways i don't care about the value i only care about its sign that's going to be positive so 2 divided by small positive of course we take to be infinity now for the second vertical asymptote so uh or the second potential location so the first in this first limit i have indeed shown that this is a vertical asymptote this function has a vertical asymptote at x equals one so now we're going to approach negative 3 halves from some direction right negative 3 halves is negative 1.5 so negative 1.5 plus 1. it's going to be negative one half in the numerator so we have a negative one half here and now i need to figure out do i have small positive or small negative so approaching negative three halves from the right so a value larger um a value larger than negative three halves so zero would certainly make our life simple here so we're approaching negative three halves from the right our test point is zero obviously plugging in zero here that's going to be zero plus zero minus one so we get a negative so i have a small negative and so negative one half divided by a small negative we take to be positive infinity and so x equals negative three halves is also a vertical asymptote so this function has two vertical asymptotes at x equals negative three halves and x equals positive one in this next example let's prove that x equals pi over two is a vertical asymptote for the tangent function in other words what i want to show is that the limit as x goes to pi over 2 of the tangent function is some form of infinity right so that's the objective here so let's look at it this way we have the limit as x goes to pi over 2. let's just pick the direction again my habit i just i'm in the habit of the right hand direction you could also pick the left-hand direction equivalently we can write the tangent as the sine of x divided by the cosine of x all right so direct substitution sine of pi over 2 is 1 cosine pi over 2 is 0. all right so our limit here plugging in pi over 2 on the top that's going to be a 1. now on the bottom we've got to figure out do are we approaching zero from the negative direction or the positive direction all right so let's approach pi over two from the right hand direction so we're going to approach pi over 2 through values larger than pi over 2. so maybe very simply we pick the value of pi as our test point right pi is certainly larger than pi over 2. so we're going to look at the cosine of pi the cosine of pi is negative 1. so we're going to be approaching 0 through the negatives so this is one divided by small negative which of course we take to be infinity all right so we've now shown that x equals pi over 2 is a vertical asymptote all right x equals pi over 2 is a vertical asymptote for the tangent function and please allow me to back up and correct myself i meant to say negative infinity in any event our conclusion is the same x equals pi over 2 is a vertical asymptote for the tangent function