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Math Help - Limits Question...

  1. #1
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    Limits Question...

    Can somebody help me understand this question better?

    lim (e^x) cos x (I only know we have to multiply all three together...)
    x->0
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  2. #2
    o_O
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    \lim_{x \to 0} e^{x} \cos x

    Because this function is continuous everywhere, just simply plug in 0 and evaluate it.
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  3. #3
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    Quote Originally Posted by o_O View Post
    \lim_{x \to 0} e^{x} \cos x

    Because this function is continuous everywhere, just simply plug in 0 and evaluate it.
    Thanks, o_O for answering.

    I know how to solve limits numerically for these questions (Just put a number close to 0, eg. .0001), but I would like to know how to solve them algebraically:

    1) lim ((2+x)^3 - 8)/x
    x->0

    2) lim (sin x)/(2x^2-x)
    x->0

    I think we have to multiply 2x by lim and 2x by sin x, but am not really sure if it's right or what to do after...

    3) lim (sin^2 x)/x
    x->0

    P.S. Tips or tricks for limits would be appreciated.
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  4. #4
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    Quote Originally Posted by AlphaRock View Post
    Thanks, o_O for answering.

    I know how to solve limits numerically for these questions (Just put a number close to 0, eg. .0001), but I would like to know how to solve them algebraically:

    1) lim ((2+x)^3 - 8)/x
    x->0

    Mr F says: Many approaches are possible. The least sophisticated is to expand the numerator and simplify. Then cancel the common factor. Then take the limit.

    2) lim (sin x)/(2x^2-x)
    x->0

    Mr F says: Many approaches are possible. Here's one: {\color{red} \frac{\sin x}{2x^2 - x} = \frac{\sin x}{x(2x-1)} = \frac{\sin x}{x} \, \frac{1}{2x-1}}.

    Now remember that the limit of a product is the product of the limits and recall {\color{red}\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1}.

    I think we have to multiply 2x by lim and 2x by sin x, but am not really sure if it's right or what to do after...

    3) lim (sin^2 x)/x
    x->0

    Mr F says: Consider the apporach in 2).

    P.S. Tips or tricks for limits would be appreciated.
    ..
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  5. #5
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    Hello,

    1) lim ((2+x)^3 - 8)/x
    x->0
    A more "sophisticated" approach is to remember (or know) the factoring formula for the difference of two cubes :
    a^3-b^3=(a-b)(a^2+ab+b^2)
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