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anonymous

  • 5 years ago

how to solve lim x->+infinity for [x+e^x+e^(2x) ]^(1/x). I need detail solution.Anyone can help?

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  1. anonymous
    • 5 years ago
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    Trying to solve this by "plugging in infinity" will give us an indeterminate result (infinity raised to 0), so we have to manipulate this to another form that can give an answer. Usually, an exponent of anything else besides an integer calls for taking the natural log of the limit argument, and then raising e to that result (because taking natural log and exponentiating cancels out).

  2. anonymous
    • 5 years ago
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    So, \[\lim_{x \rightarrow \infty} (x + e^x + e^{2x})^{1/x}\] \[= e ^{\lim_{x \rightarrow \infty} \ln((x+e^x+e^{2x})^{1/x}}\]

  3. anonymous
    • 5 years ago
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    A property logarithms allows us to move the (1/x) to the front of the natural log, so:

  4. anonymous
    • 5 years ago
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    \[e ^{\lim_{x \rightarrow \infty} \frac{1}{x}\ln(x + e^x + e^{2x})}\]

  5. anonymous
    • 5 years ago
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    at this point, trying to evaluate the limit will give us infinity over infinity, but at least it's in a form that allows us to use another technique, L'hopital's rule, which involves taking the limit of the derivative of the numerator, over the derivative of the denominator

  6. anonymous
    • 5 years ago
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    \[e ^{\lim_{x \rightarrow \infty} \frac{\frac{1+e^x + 2e^{2x}}{x+e^x + e^{2x}}}{1}}\]

  7. anonymous
    • 5 years ago
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    If you keep doing L'hopital's rule, it would never end, but it suffices to recognize as x grows, only the leading term has an affect on the value, so we can take the limit of [2e^(2x)][e^(2x)], which gives 2. Finally, exponentiate to get the aswer: e^2

  8. anonymous
    • 5 years ago
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    not really understand this part - take the limit of [2e^(2x)][e^(2x)], which gives 2.

  9. anonymous
    • 5 years ago
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    Oops, typo. It should read "take the limit of [2e^(2x)]/[e^(2x)], which gives 2". Does that make it clearer?

  10. anonymous
    • 5 years ago
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    Thank you! =)

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