let's talk about limits at infinity what is the limit as x approaches infinity of 1 over x what's the answer well let's start plugging in large numbers whenever the denominator is large you're going to get a small value if the denominator has a small value you get a large value so for example 1 divided by 10 is 0.1 1 divided by 100 is one and one divided by a thousand is point zero zero one therefore we could say that one over infinity which is a very very large number is going to be about zero so the limit as x approaches infinity of one over x is therefore equal to zero now what about the limit as x approaches negative infinity of one over x one divided by negative infinity will still be zero if you divide the one by negative point one i mean by negative ten rather this would be negative point one if you divide one by negative a hundred you get negative point zero zero one i mean negative point zero one and if you divide it by negative a thousand you get negative point zero zero one it's still approaching zero just from the left side so this is zero from the right side this is equal to zero from the left side which you could just say it's equal to zero now let's graph this function the graph of one over x looks like this it has a vertical asymptote at x equals zero and it has a horizontal asymptote at y equals zero so as you can see as x approaches positive infinity you're going to get the horizontal asymptote which is y is equal to zero as we follow the curve all the way to the right it's going to get closer and closer to zero this will also give you the horizontal asymptote on the left side which is also y equals zero to get the vertical asymptote as x approaches zero from the right side this will give us positive infinity so that tells you that zero is a vertical asymptote if you approach zero from the left side you're going to get negative infinity it's going down so the x value that leads to a y value of infinity that's going to be the vertical asymptote when x approaches infinity the y value corresponds to the horizontal asymptote let's try another problem what is the limit as x approaches infinity of 1 over x squared well here we have another bottom heavy function and if we plug in a large number let's say like a hundred one hundred squared is ten thousand and one divided by ten thousand is still a very small number point zero zero zero one so we can say this is going to approach 0. anytime you have 1 divided by infinity it's always going to equal 0. so here's the general theorem the limit as x approaches infinity of some function 1 over x to the r will always be equal to 0. so that's something that you may want to keep in mind what is the limit as x approaches infinity of eight divided by three x plus four so what process can we use to figure this out what we need to do is multiply the top and the bottom by one over x and so this is going to be equal to 8 divided by x and then on the bottom we're going to distribute 3x times 1 over x is 3 and 4 times 1 over x is 4 over x so now we can apply the limit as x approaches infinity for every term in this fraction so recall anytime you have let's say a fraction like 1 over x that's going to turn into zero the limit as x approaches infinity for one over x is zero so we can rewrite this as the limit well this is going to be let's separate the 8 from 1 over x so it's 8 times the limit as x approaches infinity 1 over x over 3 plus 4 times the limit as x approaches infinity 1 over x if you want to show every step so we know this portion is equal to 0 and the same is true for this part so it's going to be 8 times 0 divided by three plus four times zero so this is zero over three plus zero which is zero over three basically the whole thing is zero so anytime the function is bottom heavy notice that the degree of the denominator is greater than that of the numerator anytime it's bottom heavy it's going to equal zero now let's work on something else what is the limit as x approaches infinity of 8x minus 5 divided by 2x plus 3. so what is the answer in this case now notice that the degree of the numerator is the same as that of the denominator they're both to the first power so they're in the first degree whenever the degree of the numerator and the denominator is the same you can simply divide the coefficients so 8 divided by 2 is 4. therefore this is going to equal 4. now let's prove it let's show our work so i'm going to multiply the top and the bottom by 1 over x so therefore this is going to be the limit as x approaches infinity 8x times 1 over x is simply 8 and then we're going to have 5 times 1 over x which is 5 over x and then this is going to be 2 plus 3 over x the limit as x approaches infinity of 5 over x is going to equal 0. if you want to write it out separately you can write it out as 5 times the limit as x approaches infinity of one over x so that's five times zero which in the end is going to be zero now three over x that's also going to change into zero so eight divided by two is four so keep in mind for those of you who want to write out every step you can write it like this if you want to and then you can apply the formula since this is a rational function it's going to equal zero but in the end you're going to get the same answer positive four what is the limit as x approaches infinity of five minus seven x cubed divided by three x plus five x cubed plus nine go ahead and try this problem so notice that the degree of the numerator is three and the degree of the denominator is three that's the highest exponent on top and on the bottom so you could say that as x becomes large five is insignificant to seven x cubed because if you plug in a thousand into x a thousand to the third is a billion times seven that's seven billion five is insignificant to that so three x and nine are insignificant to five x cubed so this is approximately equal to uh negative seven x cubed divided by five x cubed which the answer turns out to be negative seven over five but let's confirm it so what we need to do is multiply the top and the bottom this time by one over x cubed since the highest degree is x cubed or 3. so this is going to be the limit as x approaches infinity 5 over x cubed minus 7 divided by 3 over x squared plus five plus nine over x cubed five divided by infinity to the third power is going to be zero anytime you have infinity in the bottom of a fraction it's zero so we're gonna have zero minus seven three over infinity squared that's going to be zero plus five plus nine over infinity cubed which is zero so in the end you get negative seven divided by five what is the limit as x approaches infinity of the square root of 16 x squared minus 8 divided by 2 x minus 5. go ahead and try it so when x is very large only the most significant term will remain or will be important 8 is insignificant and 5 is insignificant so this expression becomes equal to the square root of 16x squared divided by 2x the square root of 16 is 4 and the square root of x squared is x and 4x divided by 2x is 2. so that's going to be the answer but now let's confirm it with a step-by-step process so we're going to multiply the top and the bottom by 1 over x so on the bottom 2x times 1 over x is simply 2 and then we're going to have 5 over x now how do we put the 1 over x inside a square root so let's say if you have x root 2 and you want to move the x into the square root you need to multiply the exponent by 2. so it's going to turn into 2x squared because the square root of x squared is x so if you want to move it back in you need to double the exponent so to move in 1 over x into the radical we need to multiply the inside by 1 over x squared 16x squared times 1 over x squared that's going to be just 16. and then we're going to have 8 over x squared divided by 2 minus 5 over x so now let's apply the limit as x approaches infinity to every term it's not going to affect the 16 because that's a constant 8 over x squared that's going to turn into 0 and 5 over x that's going to be zero as well the square root of 16 is four so four divided by two is two and that's going to be the answer now let's try this one what is the limit as x approaches infinity of the square root of nine x to the sixth minus x squared divided by three x to the third plus one so we know that these terms are insignificant so this becomes equal to the square root of nine x to the sixth over three x cubed the square root of nine is three the square root of x to the sixth is three x to the third so these two cancel the final answer is one now to show your work we can multiply the top and the bottom by one over x cubed so this becomes the limit as x approaches infinity and then on the inside we're going to have 1 over x cubed actually x to the 6 now the exponent will double once we move it to the inside so 1 over x to the 6 times 9 x to the 6 minus x squared divided by 3 plus 1 divided by x cubed so now what we're going to do is rewrite this expression so this is going to be equal to the square root of 9 plus the limit as x approaches infinity and that's going to be negative 1 over x to the fourth that's negative x squared divided by x to the sixth and then on the bottom we're gonna have three plus the limit as x approaches infinity one over x cubed so this is going to become the square root of nine and the limit as x approaches infinity for one over x to the fourth that's going to be zero and for one over x cubed that's going to be zero and then the square root of nine is three so three divided by three is one what is the limit as x approaches infinity of this function the square root of nine x squared plus x minus three x what do we need to do here what should we do well if you're not sure what to do use direct substitution plug in a very large number into the function so let's use a thousand so nine times a thousand squared plus a thousand minus three times a thousand now if you type this whole thing in you should get point one six six six six two zero three seven three which is basically one over six which is point one six repeating so one over six is the exact answer but now what techniques can we use to actually get that answer the first thing you should do is write this as a fraction put it over one and then multiply the top and the bottom by the conjugate that is nine x squared plus x plus three x so when we foil the binomials on top the two middle terms will cancel so what we're going to have is the limit as x approaches infinity of 9x squared plus x the square roots cancel and then 3x times 3x is 9x squared but that's going to be negative 9x squared divided by what we have here these two add up to nine so we can get rid of it so based on what remains we're going to multiply the top and the bottom by one over x so this is going to be the limit as x approaches infinity x times one over x is one now inside the radical we need to multiply it by one over x squared nine x squared times one over x squared is going to be nine x times one over x squared is going to be one over x and three x times one over x is just three so now all we do all we need to do excuse me is apply the limit so one over x the limit as x approaches infinity of one over x is zero so this is going to be one divided by the square root of nine plus zero plus three and we know the square root of nine is three and 3 plus 3 is 6. so the final answer as you mentioned before is 1 over 6. what is the limit as x approaches infinity of the function arc tangent of x so what is the arc tan of infinity well let's plug in points so first put your calculator in degree mode so if we plug in arc tan of a hundred this is 89.4 now if we plug in arc tan of a thousand this is 89.94 notice that it gets closer and closer to 90. so this is equal to 90 degrees or in radians pi over 2. if we recall the limit as tangent or as x approaches pi over 2 from the left side of tangent x that's equal to infinity so when dealing with inverse tangent you need to switch these two values now what is the limit as x approaches negative infinity of the arctangent function so if we plug in arc tan of negative 1000 while it's in degree mode this is going to be negative 89.9 so this is negative pi over 2 or basically negative 90. tangent is negative in quadrants two and four but arctan only exists in quadrants one and four so that's why we got the answer that's in quadrant four negative 89.9 which gets it very close to negative 90 or negative pi over 2. so make sure you know these values now there's one more that we need to go over what is the limit as x approaches infinity of e to the negative x what is the answer so if we plug in infinity this is going to be e to negative infinity which is 1 over e to the infinity e to the infinity is infinity and one divided by infinity is zero so as x approaches infinity for this graph it's going to be zero so even if we have this one too the limit as x approaches negative infinity for e to the positive x this would be e to negative infinity which is basically 1 over infinity that 2 is 0. if you were to graph e to the x it looks like this notice that it has a horizontal asymptote of y equals zero so as x approaches negative infinity that means if we follow a curve all the way to the left the y value becomes zero