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.