anonymous
  • anonymous
how to solve lim x->+infinity for [x+e^x+e^(2x) ]^(1/x). I need detail solution.Anyone can help?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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).
anonymous
  • anonymous
So, \[\lim_{x \rightarrow \infty} (x + e^x + e^{2x})^{1/x}\] \[= e ^{\lim_{x \rightarrow \infty} \ln((x+e^x+e^{2x})^{1/x}}\]
anonymous
  • anonymous
A property logarithms allows us to move the (1/x) to the front of the natural log, so:

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anonymous
  • anonymous
\[e ^{\lim_{x \rightarrow \infty} \frac{1}{x}\ln(x + e^x + e^{2x})}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
\[e ^{\lim_{x \rightarrow \infty} \frac{\frac{1+e^x + 2e^{2x}}{x+e^x + e^{2x}}}{1}}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
not really understand this part - take the limit of [2e^(2x)][e^(2x)], which gives 2.
anonymous
  • anonymous
Oops, typo. It should read "take the limit of [2e^(2x)]/[e^(2x)], which gives 2". Does that make it clearer?
anonymous
  • anonymous
Thank you! =)

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