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Homework Help: Trig limit

  1. Sep 11, 2009 #1
    I know this probably pre-calc, but this was assinged to us in our calc class.

    1. The problem statement, all variables and given/known data
    Find the lim as h approaches zero of [tex] x * cos x [/tex]

    3. The attempt at a solution

    [tex] \frac{(x+h)cos(x+h)-(x cos x)}{h} [/tex]

    [tex] \frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h} [/tex]

    [tex] \frac{h(cos (x) cos (h) + sin (x) sin (h)) -cos (x)}{h} [/tex]

    Don't know what to do next.
  2. jcsd
  3. Sep 11, 2009 #2
    Assuming you were seeking the derivative, your parentheses are off and there is no distribution in the second term of the numerator (it's a product not a sum). Your second line should read:

    [tex]= \frac{(x+h)(cos (x) cos (h) + sin (x) sin (h)) -(x \cdot cos (x))}{h} [/tex]

    HOWEVER you state you are being asked to find the limit of xcos(x) as h approaches zero. I'm not sure you have gotten the question correct. xcos(x) does not contain an "h" and finding a limit is not equivalent to finding a derivative. Please state the question exactly as it appears so we can help you better.

  4. Sep 11, 2009 #3
    Oh, sorry, we were supposed to use the definition of a derivative.

    [tex] \frac {f(x+h)-f(x)}{h} [/tex]

    Where [tex] f(x) = x * cos(x) [/tex]
  5. Sep 11, 2009 #4


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    Homework Helper

    When taking limits, or solving maths problems in general. One should first determine what / which terms are problematic, then, isolate them, and finally, think of a way to get rid of them; instead of just expanding everything out without any goals, or reasons, and get a huge messy bunch.

    This step is bad, don't expand it early like that.

    So, our limit is:

    [tex]\lim_{h \rightarrow 0} \frac{(x + h) \cos (x + h) - x\cos x}{h}[/tex]
    [tex]= \lim_{h \rightarrow 0} \frac{x \cos (x + h) + h\cos(x + h) - x\cos x}{h}[/tex]

    Now, look at the expression closely, which terms will produce the Indeterminate Form 0/0?

    [tex]= \lim_{h \rightarrow 0} \frac{\color{red}{x \cos (x + h)} \color{blue}{+ h\cos(x + h)} \color{red}{- x\cos x}}{h}[/tex]

    The red ones, when simplifying will produce 0/0, right? And, when simplifying the blue term, by canceling 'h' will produce a normal term right? So, you limit now becomes:

    [tex]= \lim_{h \rightarrow 0} \frac{\color{red}{x \cos (x + h)} - \color{red}{x\cos x} \color{blue}{+ h\cos(x + h)}}{h}[/tex]

    [tex]=\lim_{h \rightarrow 0} \frac{\color{red}{x \cos (x + h)} - \color{red}{x\cos x}}{h} + \lim_{h \rightarrow 0} \frac{\color{blue}{h\cos(x + h)}}{h}[/tex]

    (isolating the problematic terms)

    [tex]=\lim_{h \rightarrow 0} \frac{\color{red}{x \cos (x + h)} - \color{red}{x\cos x}}{h} + \lim_{h \rightarrow 0} \color{blue}{\cos(x + h)}}[/tex]

    [tex]=\lim_{h \rightarrow 0} \frac{\color{red}{x ( \cos (x + h)} - \color{red}{\cos x} )}{h} + \color{blue}{\cos(x)}[/tex]

    Let's see if you can continue from here. :)


    And please review your algebraic manipulations, you make quit a lot mistakes in your first post: missing parentheses, and you even change the * operator to +.. @.@

    [tex]xy \neq x + y[/tex]. The 2 operators are totally different!!!!

    And [tex]-(xy) \neq (-x) * (-y) \neq (-x) + (-y)[/tex]

    I think you should really, really need go over algebraic manipulations again..
    Last edited: Sep 11, 2009
  6. Sep 11, 2009 #5
    Would you expand it there? I haven't worked with trig functions that closely so I'd assume that knowing the indeterminate form will come from practice. Edit: from looking at it.

    Haha yeah, I've slept ~3 hours in the past 1.5 days because I worked 3rd shift last night
    Last edited: Sep 11, 2009
  7. Sep 11, 2009 #6


    Staff: Mentor

    I don't think it has been pointed out that you are using a trig identity incorrectly. cos(a + h) = cos a cos h - sin a sin h.
  8. Sep 11, 2009 #7
    Is that why the identity said [tex]\mp[/tex] and not [tex]\pm[/tex] ? I was never really taught trig identities and functions in high school, so I pick up as I go through college.
  9. Sep 11, 2009 #8


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    Homework Helper

    I suggest taking the x in front of the limit and then taking a good look at the red part. It is the definition of...?
    Last edited: Sep 11, 2009
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