in this video we'll use Lobo rule to evaluate the limit as X goes to Infinity of 1 + 3x to X first let's begin by analyzing the form of the expression as X becomes large this term here 3 overx will go to zero the base of the exponential expression that we're seeing will go to one and then the power the exponent is going to Infinity so this is of the form 1 to the infinity I put that in quotes because this isn't a very helpful uh calculation we've done here in fact this is an indeterminate form it's far from Clear what the answer actually is on the one hand if I was just to think about the base here getting closer and closer to one I might expect its powers to be getting close to one on the other hand looking at this growing exponent makes me think that perhaps the expression will go to infinity and the truth can be anywhere in between we're going to use lobo's rule to calculate this limit I'm going to call it l and instead of calculating it directly I'm going to calculate the logarithm of L so I'm going to take Ln the natural log of L and apply that to the Limit and it turns out that doing this will turn this exponential form one to the infinity into a product and then we can get it into a form that lobatos rule will uh be applied to now because the logarithm is a continuous function I'm allowed to bring that Ln inside the log so the log of the limit is the limit of the logarithm of that expression and again that's because the log is a continuous function one of the properties of logarithms allows me to take the exponent x inside the log and bring it out as a factor so this becomes x * the log of 1 + 3x let's just consider the form of the expression in this new uh way of writing it X is going to Infinity so this this fact is going to infinity and then 1 + 3x is going to 1 and so log of 1 is going to zero and so what we see is something that now is a product Infinity time zero and that's still an indeterminate form it's not clear what's happening one factor is growing the other factor is shrinking what happens is very unclear it depends on how fast they're each approaching their individual limits to what the final answer will be lobo's rule can be applied only if we get this into a fractional form if I see 0 over 0 or Infinity over infinity and one way to make that happen here is to take that X and instead of multiplying it I'm going to divide by its reciprocal these are equivalent expressions and now what happens is this is going to zero up top and as X becomes large the denominator now is going to zero as well and I see a form of 0 over Z Lobos rule can be applied to that form let's carry that out now LOL says that I should take the derivative of the numerator and divide by the derivative of the denominator in the numerator I need to use the chain rule so it's going to be one over the expression inside the logarithm times the derivative of that expression 1 + 3x in the denominator I'm differentiating x to the minus1 so using the power rule this will be - 1x^2 let's finish the calculation the derivative of 1 + 3x just differentiating that 3x just like the denominator will be Min - 3x^2 and in fact this expression here in the brackets is three times as much of the expression in the denominator they will cancel leaving behind a factor of three and so doing that cancelling we'll get 1 over 1 + 3x * 3 we now reassess this situation the limit is taking X to Infinity this denominator term will go to zero so I'll be left with one or sorry three in the numerator and then 1 + 0 in the denominator and that is no longer in determinant that is equal to three keep in mind we calculated the log of L and I was interested in L itself so what we've concluded here is that the log of L is three to find L I need to exponentiate that answer e to the log of L and in this case that's e to the thir power so e cubed is the limit that we were interested in what I'm going to do is calculate the limit as X approaches zero to the right of the exact same function so nothing else has changed here we're just changing uh where we're taking the limit let's analyze the form now as X Goes To Zero from the right this term here will go to infinity and so this base now will go to Infinity while the exponent will go to zero and so the form of this is infinity to the zero that's in indeterminate form again we're going to go through the same kind of calculation I'm going to let L be the limit that I'm interested in I'm going to take the log of L that means I can take that power down that X down and I'll have x times the log let's reassess the form this factor is going to zero this Factor here will be taking the log of a larger and larger number and the logarithm is a growing function so this will go to Infinity so this is 0 * Infinity still indeterminant let's do the same trick we did a moment ago and turn it into a ratio of uh two expressions now as X goes to Infinity what's going to be or X goes to zero I'm going to see INF Infinity over infinity now and that's a situation for which lobo's rule applies we do the exact same calculation we take the derivative of the numerator and the denominator this is what we wound up with 3 over 1 + 3x same same steps that we did in the previous slide and now I let X go to zero what happens is that this term here will go to infinity and I will see a denominator that's increasing will see 3 over something getting large and this is no longer indeterminate form if I have something getting close to three the denominator increasing in size that ratio will decrease in size towards zero so the limit will be zero don't forget though that this is the log of the limit we're interested in this means the limit we really wanted L itself is e to that value so e to the zero and in this effect giving us equal to one we can conclude just by drawing the graph of this function that we were taking the limit of here is what it looks like as X goes to Infinity our first limit told us that it was going to e cubed so we're seeing an ASM toote here and as X Goes To Zero from the right we saw that it got closer and closer to e to the zero or one and we're going to see a graph that grows towards that ASM toote so here's a horizontal ASM toote and there's a removable discontin it right here noticed that the function is not defined at zero so there's really a hole in the graph there at x equal to0