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Math Help - Finding the limit

  1. #1
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    Finding the limit

    Hi, can anyone help me approach the following problems:

    Find the limit, or explain why it does not exist.

    1) lim (1-|x|) / (x-1) as x-->1
    2) lim ((sq. rt(1-h))-1) / h as h-->0

    Using a computer or calculator, estimate the following limits. Use the technology to sketch the graph of the function to check the accuracy of your estimate.

    lim (e^(2x) -1 ) / x as x-->0

    I know for this last one it says to use a calculator or computer.. but then I really find no point in even doing the problem at all if we can just use a calculator. So is there a way to solve this limit algebraically? Is there an easy way to graph this as well? Or do I just have to sub in values and graph from the table of values?

    Thanks!
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  2. #2
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    Re: Finding the limit

    For the third one, you'll have more tools for computing this limit later (L'Hospital's rule)...
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  3. #3
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    Re: Finding the limit

    Quote Originally Posted by bhuang View Post
    Hi, can anyone help me approach the following problems:

    Find the limit, or explain why it does not exist.

    1) lim (1-|x|) / (x-1) as x-->1
    2) lim ((sq. rt(1-h))-1) / h as h-->0

    Using a computer or calculator, estimate the following limits. Use the technology to sketch the graph of the function to check the accuracy of your estimate.

    lim (e^(2x) -1 ) / x as x-->0

    I know for this last one it says to use a calculator or computer.. but then I really find no point in even doing the problem at all if we can just use a calculator. So is there a way to solve this limit algebraically? Is there an easy way to graph this as well? Or do I just have to sub in values and graph from the table of values?

    Thanks!
    For the first, note that \displaystyle |x| = x \textrm{ if }x \geq 0, so since you are making the limit approach 1, \displaystyle 1 - |x| = 1 - x = -(x - 1).

    Can you go from here?

    For the second, try multiplying numerator and denominator by the numerator's conjugate...
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  4. #4
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    Re: Finding the limit

    with regards to question number 3, sort of:

    if you are willing to accept that:

    e^x = 1 + x + x^2/2 + x^3/6 + ....+ x^n/n! +....

    (which you probably have not demonstrated yet, but it is true) then,

    e^{2x} = 1 + 2x + 4x^2/2 + 8x^3/6 +....+ 2^nx^n/n! +....

    and e^{2x} - 1 = 2x + 4x^2/2 + 8x^3/6 +....+ 2^nx^n/n! +....

    so \frac{e^{2x}-1}{x} = 2 + x(2x + 4x^3/3 +....+2^nx^{n-1}/n!+....)

    if, further, you are willing to accept that the stuff in the parentheses is finite (even though it has an infinite number of terms) then:

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

    now, that's a lot of stuff to take on faith, and so its appropriate to just make a guess based on how it behaves. my suggestion is that you guess "2", and remember this limit, so you can check later, when you have more theorems under your belt.
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  5. #5
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    Re: Finding the limit

    For (2), \lim_{h\to 0} \frac{\sqrt{1- h}- 1}{h}, multiply both numerator and denominator by \sqrt{1- h}+ 1.
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