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Math Help - Can someone help me?

  1. #1
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    Question Can someone help me?

    lim [(e^3x)-1]/5x = [(e^0)-1]/5(0)=1-1/0=0/0, so the limit does not exist
    x-->0

    Is the conclusion that was drawn correct?If,so explain why.If not why not.If you believe the limit in the question exists,then propose a value for the limit and support your answer.
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  2. #2
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    Quote Originally Posted by mathcalculushelp View Post
    lim [(e^3x)-1]/5x = [(e^0)-1]/5(0)=1-1/0=0/0, so the limit does not exist
    x-->0

    Is the conclusion that was drawn correct?If,so explain why.If not why not.If you believe the limit in the question exists,then propose a value for the limit and support your answer.
    You must be careful with that line of reasoning here is an example.


    \lim_{x \to 1}\frac{x-1}{x^2-1}

    using your logic above when we evaluate at x=1 you get

    \lim_{x \to 1}\frac{x-1}{x^2-1}=\frac{1-1}{1-1}=\frac{0}{0} is undefinded

    but that is not correct now factor the denominator and reduce to get

    \lim_{x \to 1}\frac{x-1}{x^2-1}=\lim_{x \to 1}\frac{x-1}{(x-1)(x+1)}=\lim_{x \to 1}\frac{1}{x+1}=\frac{1}{2}

    To convince your self draw a plot of f(x)=\frac{x-1}{x^2-1}

    Then graph the function g(x)=\frac{e^{3x}-1}{5x} and guess what the limit will be.

    This is called an indeterminate form i.e \frac{0}{0} or \frac{\infty}{\infty} both types of limits can be calculated using L'Hosiptials rule or better yet infinite series.
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