lim x-0 (1+x)^lnx

[thanh loan]

please help me
lim x-0 (1+x)^lnx:-D

 

[Benjamin]

Re

please help me
lim x-0 (1+x)^lnx:-D

Let me help you.

(1 + x)^lnx = [e^(ln(1 + x))]^{ln x} = e^[(ln x)ln(1 + x)].

As x -> 0, ln(1 + x) behaves like x (remember the Taylor expansion
of ln (1 + x) about the point x = 0?). Thus,

(1 + x)^(ln x) behaves like e^[(ln x) x] as x -> 0.

And, as x -> 0, (ln x)*x approaches... what? You know that.

The rest is much easier and so left for ya.

Benjamin

 

[thanh loan]

thank you so muchhhh!!! ))))) i see

thank you!!

Let me help you.

(1 + x)^lnx = [e^(ln(1 + x))]^{ln x} = e^[(ln x)ln(1 + x)].

As x -> 0, ln(1 + x) behaves like x (remember the Taylor expansion
of ln (1 + x) about the point x = 0?). Thus,

(1 + x)^(ln x) behaves like e^[(ln x) x] as x -> 0.

And, as x -> 0, (ln x)*x approaches... what? You know that.

The rest is much easier and so left for ya.

Benjamin

 

[DrPhil]

Let me help you.

(1 + x)^lnx = [e^(ln(1 + x))]^{ln x} = e^[(ln x)ln(1 + x)].

As x -> 0, ln(1 + x) behaves like x (remember the Taylor expansion
of ln (1 + x) about the point x = 0?). Thus,

(1 + x)^(ln x) behaves like e^[(ln x) x] as x -> 0.

And, as x -> 0, (ln x)*x approaches... what? You know that.

The rest is much easier and so left for ya.

Benjamin

Excellent!

I think it can be written more simply, using the relationship \(\displaystyle \log(a^b) = b\ \log(a)\)

\(\displaystyle X = \displaystyle \lim_{x\to 0}\ (1+x)^{\ln x}\)

Since a variable occurs in the exponent, take the logarithm of the limit = the limit of the logarithm

\(\displaystyle \displaystyle \ln X = \lim_{x\to 0}\ \left[ \ln(1+x)\times \ln x \right]\)

Then apply the Taylor series for \(\displaystyle \ln(1 + x) = x - \ \cdot\ \cdot\ \cdot\), keeping only the first term

\(\displaystyle \displaystyle \ln X = \lim_{x\to 0}\ \left( x\times \ln x \right) = \lim_{x\to 0}\ \left(\dfrac{\ln x}{1/x}\right)\)

After you have found the limit for \(\displaystyle \ln X\), your final answer is \(\displaystyle \displaystyle X = e^{ln X}\).

 

[thanh loan]

thanks

thank all guys! )

 

[lookagain]

please help me
lim x-0 (1+x)^lnx:-D

thanh loan, you had better amend the problem to the following, because ln(x) isn't defined for x < 0 (as well as x = 0):


\(\displaystyle \displaystyle\lim_{x\to\ 0^+} \ (1 + x)^{ln(x)}\)