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Limits and Derivatives

  1. Aug 27, 2006 #1
    So we are learning about limits right now and I can't do my homework because I don't understand. Is there anyone who can explain to me how you do things such as estimate the value of ..., explain why a quantity does or does not exist and how you go about using a graph to get all your answers.

    Thanks,

    sugarcoatit
     
  2. jcsd
  3. Aug 27, 2006 #2
    I need more information to help you out. What specific problem are you having problems understanding? If you need a general explanation/definition of a limit, I suggest looking at your textbook, on Wikipedia, or asking a teacher.
     
  4. Aug 27, 2006 #3
    To estimate the value of a limit, let’s say lim f(x) as x->c you guess what f(x) gets really really close too as x gets closer to c. For example let’s say we want to estimate the limit of f(x) = (x^2-1)/(x-1) as x->1. Let’s first notice that f(x) is actually undefined at 1, since the denominator would be 0. But we can still talk about what happens as we get close, so let’s pick some trial numbers that progressively get closer to 1. how about 2,1.5,1.1, 1.01, and 1.0001.

    So
    f(2) = 3
    f(1.5) = 2.5
    f(1.1) = 2.1
    f(1.01) = 2.01
    f(1.0001) = 2.0001

    It seems as x gets closer to 1, f(x) gets closer to 2. If we do the same thing from the other side (x values less than 1) it will get close to 2 also. So a good estimation of our limit would be 2. But notice this is by no means an exact or even good process, soon you will learn a formal way to tell exactly what limits are equal too.

    For your next question. When does a limit exist? Since I don’t think you have the formal definition of a limit yet I’ll give you an intuitive one. The limit of f(x) as x-> c exist if as x gets arbitrarily close to c then f(x) gets arbitrarily close to some unique value. So let’s take f(x) = 1/x as x-> 0. This limit doesn’t exist because f(x) at 0 doesn’t get close to a value let alone a unique value. In this case we say f(x) diverges at c. Let’s consider another case f(x) = x+1 if x =< 2, x^2 if x > 2. Lim f(x) as x->2 doesn’t exist. Because x+1 gets close to 3, and x^2 gets close to 4. So as x gets close to 2, f(x) doesn’t approach a single value, it approaches 2.
     
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