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Homework Help: More Limits

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the values of a,b & c so that
    [tex]\lim_{x→0} \frac{ae^x-bcos(x)+ce^{-x}}{x\cdot sin(x)}=2[/tex]

    2. Relevant equations

    3. The attempt at a solution
    The only thing which came to my mind is that when we put x=0, it should be a 0/0 form since the limit has a finite value. Therefore, a-b+c=0.
    I can't think of any other equation to find the values of a,b and c.
    Last edited: Jun 2, 2012
  2. jcsd
  3. Jun 2, 2012 #2
    Hi again!! :wink:

    You have to use the L'Hospital rule for this. Applying the rule once will give you the second relation between a,b,c. L'Hospital'ing it another time will give you yet another relation. 3 equations, 3 variables....
  4. Jun 2, 2012 #3
    Lol, i forgot to mention, i am not allowed to use L'Hospital rule here.
  5. Jun 2, 2012 #4
    Bah, this is the seventh question I'm trying today without L'Hospital rule. Poor guy.
  6. Jun 2, 2012 #5
    I too feel so poor when i am not allowed to use L'Hospital rule. This is the seventh question for you, i have been spending my whole day to solve the Limits question without L'Hospital rule. At last, when i get exhausted, i come to PF so as to get some help! :smile:
  7. Jun 2, 2012 #6
    Oh, ouch! Take a break? :smile:

    I was pitying the person who made the rule, actually. No one bothers to use it :frown:

    Anyway, back on topic.....(trying...)
  8. Jun 2, 2012 #7


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    Can you use Taylor expansion?

  9. Jun 2, 2012 #8
    We aren't yet taught this.
    I know a very little about Taylor expansion but never used it.
  10. Jun 2, 2012 #9
    Are there any limits you know/can use with ex and e-x?
  11. Jun 2, 2012 #10
    The one in my mind is [itex]\lim_{x\to 0} \frac{e^x-1}{x}=1[/itex].
  12. Jun 2, 2012 #11


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    Well, you *have* to be able to use *something*. Otherwise, the problem can't be solved.

    If you've been taught Taylor expansion, you should use it. Look up the series expansions for [itex]e^x[/itex] (from which you can immediately deduce that for [itex]e^{-x}[/itex], [itex]\sin x[/itex] and [itex]\cos x[/itex].

    You'll need the actual terms up to and including [itex]x^2[/itex] since the denominator is [itex]x\sin x[/itex]. The remaining terms can be grouped together as [itex]O(x^3)[/itex] for ease.

    Separate out the terms and figure out what each rational expression needs to be at the limit.
  13. Jun 2, 2012 #12
    No, we aren't taught Taylor expansion yet but i do know the series you are asking me.
    Even if i expand and write [itex]e^{-x}[/itex], [itex]e^{-x}[/itex], [itex]\sin x[/itex] and [itex]\cos x[/itex], i can't take out the factor x^2 from the numerator.
  14. Jun 2, 2012 #13


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    From that you can find [itex]\displaystyle \lim_{x\to 0} \frac{e^{-x}-1}{x}[/itex]

    Rewrite your original expression as:

    [itex]\displaystyle \lim_{x→0} \frac{a(e^x-1)+a-(a+c)\cos(x)+c(e^{-x}-1)+c}{x\sin(x)}[/itex]

    Then, the identity, [itex]\displaystyle 1-\cos(x)=2\sin^2\left(\frac{x}{2}\right)[/itex] may get you nearly home.
  15. Jun 2, 2012 #14
    Thanks SammyS for the reply! :smile:
    Rewriting the expression you gave me,
    [tex]\lim_{x\to 0} \frac{a(e^x-1)+c(e^{-x}-1)+(a+c)(1-cos(x))}{x\sin(x)}[/tex]
    I get only one equation a-c=0 or a=c, i still need one more equation.

    EDIT: Thanks SammyS, i have got it! Thank you all for the help! :)
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