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Mathematics 21 Online
OpenStudy (anonymous):

Please help me this, estimate the value of the limit lim x->0 (1+x)^1/x

OpenStudy (anonymous):

\[\lim_{x \rightarrow 0} (1+x)^{1/x}\] is the answer 0?

geerky42 (geerky42):

We can let \(\large L = (1+x)^{1/x} = e^{\ln L}\), So \(\Large \displaystyle \lim_{x\rightarrow0}L = \lim_{x\rightarrow0}e^{\ln L} = \lim_{x\rightarrow0}e^{\frac{\ln(1+x)}{x}} =e^{\lim_{x\rightarrow 0}\frac{\ln(1+x)}{x}}\) After direct substitution, you will get indeterminate form: 0/0, so apply L'Hopital's rule. I'm sure you can handle it on your own afterward. Hope this helps!

geerky42 (geerky42):

Does this help? Need me to clarify more? @gogonhan

OpenStudy (anonymous):

can you please show more work, i dont get it... Sorry I just got Caculous this morning... so a lot of confusing...

geerky42 (geerky42):

which part are you confused with?

OpenStudy (anonymous):

okie... I got it... just read the book... thank you :D

geerky42 (geerky42):

no problem!

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