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Limits with trig functions

  1. Feb 13, 2006 #1
    I'm pretty useless at trig manipulations, but I'm trying to learn

    According to my text, the key to limits with trig functions is to get them into this form

    [tex] \lim_ {t\rightarrow 0} \frac{\sin x}{x} = 1 [/tex]

    I've been doing alright with that, but I'm stuck on this one
    [tex] \lim_ {t\rightarrow 0} \frac{\tan 6t}{\sin 2t} [/tex]

    I'm not sure what else to turn this into besides
    [tex] \lim_ {t\rightarrow 0} \frac{\sin 6t}{\sin 2t \cos 6t} [/tex]

    A poke in the right direction would be appreciated.

    Thanks
    Jeff
     
    Last edited: Feb 13, 2006
  2. jcsd
  3. Feb 13, 2006 #2

    quasar987

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    You could use the identity

    [tex]tg2x=2tgx/(1-tg^2x)[/tex]

    I tought I read in another thread that you teach a calculus class. Did I misread?
     
  4. Feb 13, 2006 #3

    HallsofIvy

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    Take a look at
    [tex]\frac{n}{m}\frac{sin 6t}{nt}\frac{mt}{sin 2t}\frac{1}{cos 6t}[/tex]

    I'll let you do the work of deciding how to choose n and m correctly.
     
  5. Feb 13, 2006 #4
    I do not teach a calculus class, I take a calculus class :tongue:
     
  6. Feb 13, 2006 #5
    Ok, I've made some progress

    [tex] \lim_ {t\rightarrow 0} \frac{\sin 6t}{\sin 2t \cos 6t} (\frac{12t}{12t}) = \lim_ {t\rightarrow 0} \frac{6}{2} (\frac{\sin 6t}{6t}) (\frac {2t}{\sin 2t}) (\frac{1}{\cos 6t})[/tex]

    I know the limit here for each term except [tex] (\frac {2t}{\sin 2t}) [/tex]

    Straight substitution leaves this undeffined. Is there a standard for working out limits in this form?
     
    Last edited: Feb 13, 2006
  7. Feb 13, 2006 #6
    Is your limit [tex]x\rightarrow 0[/tex] or [tex]t\rightarrow 0[/tex]? Anyway, I seem to remember that even
    [tex]\lim_{x\rightarrow 0} \frac{x}{\sin{x}}=1[/tex]
    but I am far from certain.
     
  8. Feb 13, 2006 #7
    It's [itex] t\rightarrow 0 [/itex]. I've made the change in the earlier posts.

    It would seem that if [tex] \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1 [/tex] then [tex] \lim_{x\rightarrow 0} \frac{x}{\sin x} = 1 [/tex], but I'm wary of making that assumption. Can someone clarify if I am correct?
    If so, the answer to my original question is 3, which matches up with the back of the text.


    Thanks

    Jeff
     
  9. Feb 13, 2006 #8
    I seem to remember this being the case. However I was not certain so I did a quick check in Mathematica and it gives me the same answers. The question is; how reliable is Mathematica? :smile:
     
  10. Feb 13, 2006 #9

    arildno

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    Do you agree that your wariness hinges upon the following:
    Given that lim f =L (non-zero), then lim 1/f =1/L ?

    What do you know about this statement?
     
  11. Feb 13, 2006 #10
    It appears to be true. I can't think of a counter example.
     
  12. Feb 13, 2006 #11

    arildno

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    That's not good enough! You should prove it!
    Here's one way:
    1. assume that the limit of f at x=0 is L (non-zero)
    Since L is non-zero, there must be a non-zero neighbourhood about x=0 where f is always non-zero (otherwise, L could not be f's limit value at x=0, since at arbitrarily close points, there would be function values differing from L at least of magnitude L)

    2. Regard the difference (within the domain existing by virtue of 1.):
    [tex]|\frac{1}{f}-\frac{1}{L}|=\frac{1}{|f||L|}|f-L|[/tex]

    try to concoct an epsilon-delta proof to show that the statement holds.
     
  13. Feb 13, 2006 #12
    I'm still in Calc I and haven't done much more than the most basic epsilon-delta proofs. I'd love to learn how, but I'm uncertain as to where to begin.

    Do I start with [tex]|\frac{1}{f}-\frac{1}{L}| < \epsilon [/tex] and [tex] \frac{1}{|f||L|}|f-L| <\epsilon [/tex] and try to find [tex] \delta > |x| [/tex] that works for both cases?
     
  14. Feb 13, 2006 #13

    arildno

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    The first thing to remember about the epsilon-delta business is not to bother overmuch about the epsilon!
    What IS in general important, is to bound your expression with a function of delta that is easily seen to go to zero as delta goes to zero.

    Now, in this case there are two salient features:
    1. We need to estimate the |f| in the denominator in a clever manner.
    2. We have the difference |f-L| to take care of.

    1. We want to find an upper bound for [itex]\frac{1}{|f||L|}|f-L|[/itex]
    Hence, we should first find a LOWER bound for |f|, since it appears in the denominator.
    Since we know that the limit of f is L, there will exist a [itex]\delta_{1}[/itex] so that for all x's inside that delta,[tex]|f|\geq\frac{|L|}{2}[/tex]
    (the half is chosen for simplicity, the validity of the inequality is straightforward to see).

    2. Since f goes to L, then for any [itex]\epsilon[/tex] there will exist a [itex]\delta_{2}[/itex] so that [itex]|f-L|\leq\frac{L^{2}\epsilon}{2}[/tex]
    (This particular choice is just cosmetics, in order to end up with a nice expression.)

    3. Thus, choosing [itex]\delta=minimum(\delta_{1},\delta_{2})[/itex]
    we get, for x's inside the delta-band:
    [tex]\frac{1}{|L||f|}|f-L|<\frac{2}{L^{2}}\frac{L^{2}\epsilon}{2}=\epsilon[/tex]
    Voila!
     
  15. Feb 13, 2006 #14
    Clever logic. It took me awhile, but I think I understand what you did. Just for reference, is this something I should be learning at the Calc I level, or does this usually come up in something like an Analysis class?
     
  16. Feb 13, 2006 #15
    Since you are in Calc 1 you can use L'Hospital's Rule to show that the limit exists and is also 1.
     
  17. Feb 13, 2006 #16

    arildno

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    And how do you prove in the first place that we have:
    [tex]\frac{d}{dx}\sin(x)=\cos(x)[/tex] ?
     
  18. Feb 13, 2006 #17
    That's about a chapter ahead of where I am now. I read a few sections ahead of the class, but not quite that far.
     
  19. Feb 13, 2006 #18
    Ah, I didn't realize you hadn't covered derivatives yet.
     
  20. Feb 13, 2006 #19
    I can follow the proof of this one almost all the way to the end, staring from the limit definition of the derivative. I just get a little stuck at the last step. I can break it down to terms that are almost all defined when [itex] h = 0 [/itex] but I get two terms in that step that are [tex] \lim_{h\rightarrow 0} \frac{\sin(h)}{h} [/tex]

    I know this equals one, but I'm not sure how to prove this logically. The proof in my text is only done geometrically.
     
  21. Feb 13, 2006 #20

    TD

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    Without using derivatives, you can show this since for small values of x (we're letting x going to 0), we have that:

    [tex]
    \begin{array}{l}
    \sin x \le x \le \tan x \\
    \frac{{\sin x}}{{\sin x}} \le \frac{x}{{\sin x}} \le \frac{{\tan x}}{{\sin x}} \\
    1 \le \frac{x}{{\sin x}} \le \frac{1}{{\cos x}} \\
    \end{array}
    [/tex]

    And since [itex]\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x}} = 1[/itex], we've now 'squeezed' our function between two which have limit 1.
     
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