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Limits problem

  1. Nov 2, 2009 #1
    I would like to discuss my solution to a particular limits problem on my exam. I was to evaluate the following limit: [tex]= \lim_{x\to 0}\frac{tan(5x)}{sin(2x)}[/tex]

    3. The following is the exact solution I wrote on the exam
    My work is in BLACK
    My professors corrections are in RED


    [tex]= \lim_{x\to 0}\frac{tan(5x)}{sin(2x)}[/tex]

    [tex]= \lim_{x\to 0}\frac{\frac{sin(5x)}{cos(5x)}}{sin(2x)}[/tex]

    [tex]= \lim_{x\to 0}\frac{sin(5x)}{sin(2x)cos(5x)}[/tex]

    [tex]= (\lim_{x\to 0}\frac{sin(5x)}{1}) (\lim_{x\to 0}\frac{1}{sin(2x)}) (\lim_{x\to 0}\frac{1}{cos(5x)})[/tex] --> the middle limit does not exist

    (rule:limit of the product is the product of the limit) --> only if the correct limits exist!!

    [tex]= (\lim_{x\to 0}\frac{5sin(5x)}{5}) (\lim_{x\to 0}\frac{1}{\frac{sin(2x)}{1}}) (\lim_{x\to 0}\frac{1}{\frac{cos(5x)}{1}})[/tex]

    [tex]= (5)(1) (\lim_{x\to 0}\frac{1}{\frac{2sin(2x)}{2}}) (\frac{1}{\frac{1}{1}})[/tex]

    [tex]= (5)(1) (\frac{1}{2}) (1)[/tex]

    [tex]= \frac{5}{2}[/tex] ??

    ----------------End of work--------------------

    So the professor gave me a score of 3/6 on this question, and I believe I deserve more. I don't know where she's going when she wrote that the middle limit does not exist, because this solution is similar to a question she solved herself in class. My solution makes sense to me, and I did get the correct answer. Does anyone else agree with me? I would like to confront the prof and argue this question. Do I have any basis for this argument, or do I really deserve 3/6?

    Thanks.
     
  2. jcsd
  3. Nov 2, 2009 #2
    Certainly you got the correct answer, but the professor is correct that the middle limit does not exist (check from both sides as approaching 0). I don't think it will be productive to argue subjective points with the professor.
     
  4. Nov 2, 2009 #3
    Then what must I have instead of the middle limit?

    Besides, I don't see why 3 mark are deducted. She was very fair in other questions, so it just seems really odd that 3 marks are deducted for making an error in one step.
     
  5. Nov 2, 2009 #4
    I think you were given half credit because you failed to see the whole "trick" while not introducing any "evil" like the professor pointed out. I don't know exactly what you know at this point in the course, but it possibly involves multiplying the limit by [tex]\frac{10x}{10x}[/tex] to yield:

    [tex]= (\lim_{x\to 0}\frac{5sin(5x)}{5x}) (\lim_{x\to 0}\frac{2x}{2sin(2x)}) (\lim_{x\to 0}\frac{1}{cos(5x)})[/tex]

    edit: whoops minor mistake.
     
  6. Nov 2, 2009 #5
    So how is your middle limit different than mine? Does it not also exist?

    You see, all I did was separate [tex]= \lim_{x\to 0}\frac{sin(5x)}{sin(2x)cos(5x)}[/tex] into 3 limits.
     
  7. Nov 2, 2009 #6
    Mine does exist. The reason why yours doesn't is that if you observe the value from infinity to 0 the limit tends toward infinity, whereas if you observe the value from -infinity to 0 the limit tends toward -infinity. Mine, however, does not, as the addition of x to the numerator causes the signs between the numerator and denominator to always cancel and therefore the limit exists as it converges to the same value from either side.
     
  8. Nov 2, 2009 #7
    So writing [tex]\lim_{x\to 0}\frac{1}{sin(2x)}[/tex] alone is incorrect?
     
  9. Nov 2, 2009 #8
  10. Nov 2, 2009 #9
    Yes I see how the prof is correct in this now, but I still don't believe my errors account for half of the solution.

    Does anybody else have something to add?
     
    Last edited: Nov 2, 2009
  11. Nov 2, 2009 #10
    Well you can take it up with the professor, but definitely keep in mind that a more craftily-made problem could have caused your thinking to give the incorrect solution. For example, the limit of [tex]\lim_{x \to 0}\frac{sin(x)}{sin(2x)sin(3x)}[/tex]
     
  12. Nov 2, 2009 #11
    Also:
    when you take lim(x-0) of 5sin(5x)/5 you get 0. not 5 as you stated. this is due to the fact that sin(ax)=0. Furthermore lim(x-0) of 1/((2sin(2x))/2) is undefined because the fraction would end up being undefined because it would be equal to 1/(0/1) when the limit is applied (again because sin(ax)=0). Then you miraculously arrive at the correct answer.
     
  13. Nov 2, 2009 #12
    So I seem to have 2 errors in my solution. But shouldn't I get at least one more mark for getting the right answer?
     
  14. Nov 2, 2009 #13
    I think the problem is that your answer comes from nowhere. You have incorrect math that somehow arrives at the correct answer. In my opinion, you should be happy with your 3/6...
     
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