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As a revision exercise we have been given.

find

[tex] \lim_{\substack{x\rightarrow infty} } (x+e^x) ^ (2/x) [/tex]

****Darn It it wont work, the above is (x+e^x) all raised to (2/x)

our lecturer has given us a walkthrough the solution which is done by letting the function = y and taking logs of both sides, which gives a limit of infinity/infinity, so we can use L'hopitals rule to derive a limit for lny to infinity then invert it onto the index of e for the answer.

This all makes sense to me, I just have one problem with the working.

the function for lny comes out to [tex] \frac {2ln(x+e^x)}{x} [/tex]

and the derivative of this is given as [tex]\frac {2(1+e^x)/(x+e^x)}{1}[/tex]

The numerator makes sense to me to be the derivative of 2ln(x+e^x) but I don't understand how the denominator becomes so simple. It looks to me like the solution just differentiates the top and then differentiates the bottom?

:uhh:

find

[tex] \lim_{\substack{x\rightarrow infty} } (x+e^x) ^ (2/x) [/tex]

****Darn It it wont work, the above is (x+e^x) all raised to (2/x)

our lecturer has given us a walkthrough the solution which is done by letting the function = y and taking logs of both sides, which gives a limit of infinity/infinity, so we can use L'hopitals rule to derive a limit for lny to infinity then invert it onto the index of e for the answer.

This all makes sense to me, I just have one problem with the working.

the function for lny comes out to [tex] \frac {2ln(x+e^x)}{x} [/tex]

and the derivative of this is given as [tex]\frac {2(1+e^x)/(x+e^x)}{1}[/tex]

The numerator makes sense to me to be the derivative of 2ln(x+e^x) but I don't understand how the denominator becomes so simple. It looks to me like the solution just differentiates the top and then differentiates the bottom?

:uhh:

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