Limits: How to approach them? The purpose of this guide is to show you the many ways to solve it problems. These depend on many factors. The best way to do this is by working out a few eamples. In particular, we will be looking at its involving piece wise functions and indefinite forms such as 0,,, 0 0 Eample 1 Given the piecewise function f() = 32 + 1, 2 ln( 1) + 1, > 2 Find the following : (a) 2 + f(), (b) 2 f(), (c) 2 f() Solution for (a): We have that 2 +, which means is getting close to 2 from the right. We are looking at numbers that are greater than 2 such as 2.01, 2.001 etc. So we have to grab the part of the function that has these numbers in its domain. That would be the part where > 2 So we will now compute the it: 2 f() = ln( 1) + 1 = ln(2 1) + 1 = ln(1) + 1 = 0 + 1 = 1 + + 2 Solution for (b): We have that 2, which means is getting close to 2 from the left. We are looking at numbers that are less than 2 such as 1.99, 1.999, etc. We have to grab the part of the function that has these numbers in its domain. That would be the part where < 2 Computing the it: f() = 2 2 32 + 1 = 3(2) 2 + 1 = 12 + 1 = 11 Solution for (c): 2 f() eists only if 2 f() = 2 + f(). But 2 + f() = 1 and 2 f() = 11. So 2 f() = does not eist Eample 2 Find 2 4 2 16 3 8 Solution: If we were to plug in 2 into our function, on both the numerator and denominator, you get the form 0. This is an indeterminate form. We will need to do some more work to know whether this it 0
eists or not. Typically, when you have the form 0, we will use algebra to simplify our function. Often, 0 this will lead you to cancel the factors that make the epression into the form 0 0 Let s go ahead and do this for the above problem. 4 2 16 2 3 8 = 2 4( 2)( + 2) ( 2)( 2 + 2 + 4) = 2 4( + 2) 2 + 2 + 4 = 16 12 = 4 3 The factor of 2 cancelled from both the numerator and denominator. This was causing the form 0 0 Limits at Infinity The following eamples deal with its at infinity i.e. or Eample 3 Find 4 3 2 2 +5 3 8+5 2 3 Solution: This it has the form which is indeterminate. Factoring is not a feasible option. So a what we will do is use the fact that ± r = 0 for any number r > 0. How will we use this fact in this problem? I am going to use algebra and divide every term on both the numerator and denominator with the highest power or highest growing term. This step will make every term look like a ± r 4 3 4 3 2 2 + 5 3 (4 3 2 2 + 5 3) 3 8 + 5 2 3 = ( 8 + 5 2 3 ) 3 = 3 22 3 + 5 3 3 3 8 3 + 52 3 3 3 4 2 + 5 2 3 3 4 8 2 + 5 = 1 2 + 5 2 3 3 = 8 2 + 5 1 4 0 + 0 0 0 0 1 = 4 The step where we divided by the highest power in both the numerator and denominator is a very useful step when dealing with infinite its. But beware! This technique only works well with infinite its and nothing else!! Eample 4 Find
2 + 4 Solution: This it has the form of. The top looks like because as goes to, 2 + 4 is positive because 2 gets larger than 4. We will approach this problem in a similar manner as we did eample 3 but with a couple differences. So what would be the highest growing power? We might want to say 2, but 2 is inside the square root. So we are really looking at just as the highest power. So let us divide by on the numerator and denominator. 2 +4 2+4 =. So what do we do with the numerator? I can t divide everything by because of the square root. What we can do is place inside the square root. We can try to rewrite as = 2. However, in reality, 2 =. But we have, not. But there is a fi for, 0 that. We can use the definition of =. Because, we will take the part that, < 0 corresponds to. Meaning, = 2 or = 2 2 + 4 = 2 + 4 2 2 + 4 = 2 = 1 2 2 + 4 2 = 1 + 4 = 1 + 0 = 1 Eample 5 Find e e e + e Solution: The reason I want to do this eample so we can review the growth of e or e. In terms of computing the it, nothing will change as the previous two eamples. As, we have that e 0 and e. (If you are unsure of why this is the true, I suggest looking at the graphs of e and e ). Once again, we have the indeterminate form of the larger of the two so I will divide every term by e e e e +e = e e e e e = e +e e e 2 1 = 0 1 = 1 e 2 +1 0+1 e will be
A couple things to note: We used some rules of algebra e e = e ( ) = e 2. Also, e a 0 if for all a > 0. Lastly, if the question had been asking for, then e would be the larger of two. The last two problems deal with the indeterminate forms and 0 Other Indeterminate forms. Eample 6 Find 2 Solution: If we were to take this it we would have the form of. Many mistaken this for 0. However, all that we are saying is our two terms grow infinitely large but may not be the same number. In other words, when 2 is 500, might be 400; when 2 is 1000, is 500. All we can say is the terms grow really large but we can t say they cancel each other. The way to handle the situation is to use algebra like we have with the previous problems. When we have functions that have a square root and a binomial inside of it, we use the conjugate a lot. The conjugate consists of multiplying both the numerator and denominator by the almost same eact binomial but with an opposite sign in the middle. 2 = ( 2 ) 2 2 + = 2 1 2 + [ 2 + ] [ 2 + ] = [ 2 ] 2 2 2 + Stopping for a moment we have the form of which is still indeterminate. However, we know how to handle the situation by dividing by the highest growing power. I will leave the reader to work out the details. The final answer will be 1 2 Eample 7 Find 1 0 + (3 3 + 1 ) If we were to look at this it by merely plugging in 0 from the right, I would have the form 0 which is indefinite. We can t say this result is going to be zero. 1 grows really large as 0+. This means can approach 500, 1000, etc. At the same time,as 0 +, (3 3 ) is approaching zero. This +1 implies that (3 3 ) can approach 1 then 1, etc. If I were to multiply these results this would mean that +1 2 4
if 1 3 is 500 and (3 ) is 1 then I would have 250. If 1 3 is 1000 and (3 ) is 1 then I would have +1 2 +1 4 250. The point that I am trying to make is that even though one part goes to and the other part goes to 0, we can t say the result will be zero. The result might be a number like 250, as the eample was showing. Again, we use algebra to solve this it: I will get a common denominator for 3 3 +1, and multiply by 1 1 0 + 3 3 = + 1 0 + 1 + 1) 3( + 1 3 = + 1 There are other indeterminate forms like 0 0, 1, 0. 0 + 1 3 3 = + 1 0 + + 1 = 3