Limit Problem

[Jason76]

\(\displaystyle \lim x \rightarrow \infty[(e^{x} + x)^{2/x} ]\) = indeterminate \(\displaystyle \infty^{0}\)

\(\displaystyle \ln[(e^{x} + x)^{2/x}\)

\(\displaystyle 2/x\ln (e^{x} + x)\)

Simplify

\(\displaystyle \dfrac{2 + \ln x}{x}\)

Use L hopitals

\(\displaystyle \dfrac{1/x}{1} = 0\)

????

\(\displaystyle \lim \rightarrow \infty[(e^{0} + 0)^{2/0} ] = 1\) ???

 

[HallsofIvy]

\(\displaystyle \lim x \rightarrow \infty[(e^{x} + x)^{2/x} ]\) = indeterminate \(\displaystyle \infty^{0}\)

\(\displaystyle \ln[(e^{x} + x)^{2/x}\)

\(\displaystyle 2/x\ln (e^{x} + x)\)

Simplify

\(\displaystyle \dfrac{2 + \ln x}{x}\)

This is your error. ln(e^x+ x) is NOT equal to ln(x)
The derivative of ln(e^x+ x) is \(\displaystyle \frac{e^x+ 1}{e^x+ x}\).
Dividing both numerator and denominator by \(\displaystyle e^x\), it is \(\displaystyle \frac{1+ e^{-x}}{1+ xe^{-x}}\).

As x goes to infinity, both \(\displaystyle e^{-x}\) and \(\displaystyle xe^{-x}\) go to 0 (you can show that by applying L'Hopital's rule to \(\displaystyle \frac{x}{e^x}\)).

Use L hopitals

\(\displaystyle \dfrac{1/x}{1} = 0\)

????

\(\displaystyle \lim \rightarrow \infty[(e^{0} + 0)^{2/0} ] = 1\) ???

 

[Jason76]

This is your error. ln(e^x+ x) is NOT equal to ln(x)
The derivative of ln(e^x+ x) is \(\displaystyle \frac{e^x+ 1}{e^x+ x}\).
Dividing both numerator and denominator by \(\displaystyle e^x\), it is \(\displaystyle \frac{1+ e^{-x}}{1+ xe^{-x}}\).

As x goes to infinity, both \(\displaystyle e^{-x}\) and \(\displaystyle xe^{-x}\) go to 0 (you can show that by applying L'Hopital's rule to \(\displaystyle \frac{x}{e^x}\)).

The solution would be to take L hopitals one more time?

 

[Jason76]

\(\displaystyle \lim x \rightarrow \infty[(e^{x} + x)^{2/x} \)

\(\displaystyle y = (e^{x} + x)^{2/x} \)

\(\displaystyle \lim x \rightarrow \infty = e^{ln y}\)

\(\displaystyle \lim x \rightarrow \infty = e^{(2/x)\ln (e^{x} + x)}\)

take L hopitals

\(\displaystyle lim x \rightarrow \infty = e^{(\dfrac{2e^{x} + 1}{e^{x} + x})/1}\)

Take l hopitals again

\(\displaystyle \lim x \rightarrow \infty = e^{\dfrac{2e^{x} }{e^{x} + 1}} = 2\)

So \(\displaystyle e^{2}\) is the answer on homework.