Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    Hey all,

    I'm working my way through the example sheets before my engineering mathematics exam on Friday and I've gotten stuck on a few limits questions, I was hoping for some help. Thanks!

    vi) lim x->0 (1-cosx)/(x^2)

    I'm not sure where to go with this, I changed cosx into (1-2sin^2(x/2)) and then manipulated it to give:

    lim x->0 (1/x) * [{sin(x/2) / (x/2)} * sin(x/2)]

    but I have no idea whether or not that's right. I know sin(x/2) / (x/2) is equal to one, and sin(x/2) is zero, but what about the 1/x outside the square brackets?

    vii) lim x-> (inverse sin x)/ x
    "Hint: let x = siny"

    I did that and then this is what I got:

    lim siny->0 (y/siny)

    and again have no idea how to take it further.

    Finally, question 5 i), ii) and iii) involve continuous and discontinuous functions. I understand that for a function be continuous at x, f(x) must be equal to f(x) approached from above and below x, but the questions don't even say what value of x to use and I'm completely and utterly confused.
    Please see the questions attached to this message.


    Thanks so much!
    Attached Images
     
    Offline

    8
    ReputationRep:
    I am rusty with limits but think in vi) you need to apply L'Hôpital's rule... This is where you use the formula  lim \frac{f(x)}{g(x)} = lim \frac{f'(x)}{g'(x)} where ' sigifies derivative. In this case  f(x) = 1 - cos x and  g(x) = x^2 . The rule can be applied ad infinitum (subject to some caveats about the derivative being defined, but this is not a problem here) so im gonna differentiate twice like this based on

    lim \frac{f(x)}{g(x)} = lim \frac{f''(x)}{g''(x)}

     f''(1 - cos x) = cos (x)

     g''(x^2) = 2

    After this I now have the far simpler  lim \frac{cos x}{2} which as x tends to 0 gives  \frac{1}{2} .

    I think same principle applies for vii) also...

    Last part - not sure about that im afraid!
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by spread_logic_not_hate)
    I am rusty with limits but think in vi) you need to apply L'Hôpital's rule... This is where you use the formula  lim \frac{f(x)}{g(x)} = lim \frac{f'(x)}{g'(x)} where ' sigifies derivative. In this case  f(x) = 1 - cos x and  g(x) = x^2 . The rule can be applied ad infinitum (subject to some caveats about the derivative being defined, but this is not a problem here) so im gonna differentiate twice like this based on

    lim \frac{f(x)}{g(x)} = lim \frac{f''(x)}{g''(x)}

     f''(1 - cos x) = cos (x)

     g''(x^2) = 2

    After this I now have the far simpler  lim \frac{cos x}{2} which as x tends to 0 gives  \frac{1}{2} .

    I think same principle applies for vii) also...

    Last part - not sure about that im afraid!
    Legend, thanks so much!! Tried vi) and vii) with L'Hôpital's rule, worked perfectly, cheers.

    Now I just need to figure out how to work question 5 haha.

    Thanks!
    Offline

    12
    ReputationRep:
    You could evaluate the limit by replacing cos x by its series expansion
    Offline

    12
    ReputationRep:
    Also, you're right about the definition of continuity at a particular point x. A continuous function is continuous at all x. So you need to check this. In practice, you can just draw these functions and see if you need to take your pen off the paper to draw them. If you do, the function is not continuous.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Drederick Tatum)
    Also, you're right about the definition of continuity at a particular point x. A continuous function is continuous at all x. So you need to check this. In practice, you can just draw these functions and see if you need to take your pen off the paper to draw them. If you do, the function is not continuous.
    Thanks but it requires the non graphical method (we can add the graph in an answer if we want to, but we have to do it via the limits method). The problem is I'm not sure what x to choose when doing this evaluation as up until now we were always told which to choose (and the questions seemed a lot easier).
    Offline

    8
    ReputationRep:
    Ok got how to do Q5 I think - basically you gotta check what the function tends to at point you're considering.

    For 5i) we have

     lim (1+x^2) at  x = 1 is  1+1^2 = 2 . Since the function is defined as  f(x) = 5 at  x = 1 then it is discontinuous, as just around  x = 1 the function does not tend to 5.

    For more info check this site http://cnx.org/content/m17442/latest/ on the types of discontinuity section.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you rather give up salt or pepper?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.