Limit Problem
- Thread starter Jason76
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\(\displaystyle \displaystyle \lim_{x \to \infty} (e^x +x)^{9/x}\)\(\displaystyle \lim x \rightarrow \infty \)
\(\displaystyle (e^{x} + x)^{\dfrac{9}{x}} \)
\(\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}} \)
\(\displaystyle (\infty + \infty)^{0} \)
\(\displaystyle (\infty)^{0}\)
Take the logarithm.
\(\displaystyle \displaystyle \ln\left(\lim_{x \to \infty} (e^x +x)^{9/x}\right) = \lim_{x \to \infty}\left(\dfrac{9\ \ln (e^x + x)}{x}\right)
\implies \ \cdot \ \cdot \ \cdot\)
It still comes out to "infinity over infinity" which is indeterminate.
What do you do with indeterminate ratios??? What method have all of these problems been about?
\(\displaystyle \lim x \rightarrow \infty \)
\(\displaystyle (e^{x} + x)^{\dfrac{9}{x}} \)
\(\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}} \)
\(\displaystyle (\infty + \infty)^{0} \)
\(\displaystyle (\infty)^{0}\) - Indeterminate
\(\displaystyle (e^{x} + x)^{\dfrac{9}{x}} \) Rewrite
\(\displaystyle \dfrac{9}{x} \ln (e^{x} + x)\)
\(\displaystyle \dfrac{9 \ln (e^{x} + x)}{x} \)
\(\displaystyle \dfrac{\dfrac{d}{dx} 9 \ln (e^{x} + x)}{\dfrac{d}{dx} x} \)
\(\displaystyle \dfrac{9 \ln (u)}{1} \)
\(\displaystyle \dfrac{\dfrac{9}{u}(e^{x} + 1)}{1} \)
\(\displaystyle \dfrac{\dfrac{9}{e^{x} + x}(e^{x} + 1)}{1} \)
\(\displaystyle \dfrac{\dfrac{9e^{x} + 9}{(e^{x} + x)}}{1} \)
\(\displaystyle \dfrac{\dfrac{9e^{\infty} + 9}{(e^{\infty} + (\infty)}}{1} \)
\(\displaystyle \dfrac{\infty}{1} \) Answer?
\(\displaystyle (e^{x} + x)^{\dfrac{9}{x}} \)
\(\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}} \)
\(\displaystyle (\infty + \infty)^{0} \)
\(\displaystyle (\infty)^{0}\) - Indeterminate
\(\displaystyle (e^{x} + x)^{\dfrac{9}{x}} \) Rewrite
\(\displaystyle \dfrac{9}{x} \ln (e^{x} + x)\)
\(\displaystyle \dfrac{9 \ln (e^{x} + x)}{x} \)
\(\displaystyle \dfrac{\dfrac{d}{dx} 9 \ln (e^{x} + x)}{\dfrac{d}{dx} x} \)
\(\displaystyle \dfrac{9 \ln (u)}{1} \)
\(\displaystyle \dfrac{\dfrac{9}{u}(e^{x} + 1)}{1} \)
\(\displaystyle \dfrac{\dfrac{9}{e^{x} + x}(e^{x} + 1)}{1} \)
\(\displaystyle \dfrac{\dfrac{9e^{x} + 9}{(e^{x} + x)}}{1} \)
\(\displaystyle \dfrac{\dfrac{9e^{\infty} + 9}{(e^{\infty} + (\infty)}}{1} \)
\(\displaystyle \dfrac{\infty}{1} \)
Answer?
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HallsofIvy
Elite Member
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- Jan 27, 2012
- Messages
- 7,763
IF it came to "\(\displaystyle \dfrac{\infty}{1}\)" then the answer would be "infinity" or "there is no limit".\(\displaystyle \lim x \rightarrow \infty \)
\(\displaystyle (e^{x} + x)^{\dfrac{9}{x}} \)
\(\displaystyle (e^{\infty} + \infty)^{\dfrac{9}{\infty}} \)
\(\displaystyle (\infty + \infty)^{0} \)
\(\displaystyle (\infty)^{0}\) - Indeterminate
\(\displaystyle (e^{x} + x)^{\dfrac{9}{x}} \) Rewrite
\(\displaystyle \dfrac{9}{x} \ln (e^{x} + x)\)
\(\displaystyle \dfrac{9 \ln (e^{x} + x)}{x} \)
\(\displaystyle \dfrac{\dfrac{d}{dx} 9 \ln (e^{x} + x)}{\dfrac{d}{dx} x} \)
\(\displaystyle \dfrac{9 \ln (u)}{1} \)
\(\displaystyle \dfrac{\dfrac{9}{u}(e^{x} + 1)}{1} \)
\(\displaystyle \dfrac{\dfrac{9}{e^{x} + x}(e^{x} + 1)}{1} \)
\(\displaystyle \dfrac{\dfrac{9e^{x} + 9}{(e^{x} + x)}}{1} \)
\(\displaystyle \dfrac{\dfrac{9e^{\infty} + 9}{(e^{\infty} + (\infty)}}{1} \)
\(\displaystyle \dfrac{\infty}{1} \)
(Saying the limit is "infinity" is just saying the limit does not exist in a particular way.)
However, it is not. The limit of \(\displaystyle \dfrac{9e^x+ 9}{e^x+ x}\) is not "infinity". But surely you know that \(\displaystyle \dfrac{\dfrac{9e^x+ 9}{e^x+ x}}{1}\) is just
\(\displaystyle \dfrac{9e^x+ 9}{e^x+ x}\)?
Divide both numerator and denominator by \(\displaystyle e^x\).
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