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Limits and also continuous/discontinuous functions watch

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    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!
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    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!
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    (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!
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    You could evaluate the limit by replacing cos x by its series expansion
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    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.
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    (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).
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    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.
 
 
 
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