lim x->infinity (1+(1/x))^x

Hi,

I have another question on limits:

lim (1+(1/x))^x

x->infinity

So i assume that I have to use L'Hopital's Rule to get to an answer

Firstly I have let y=the equation and taken the natural log of each side:

ln(y)=ln[(1/(1+(1/x)))^x]

and then made it the form:

=lim x*ln(1/(1+(1/x)))

x->infinity

=lim ln(1+(1+(1/x)))/x^-1

x->infinity

and used L'hopital to get to:

1/(1+(1/x))

Everywhere i have looked has said that the answer is that x approaches e and I have absolutely no idea how I get this solution? If you are able to give me a hand it would be greatly appreciated! Thanks, sorry it is so messy I don't know how to write it out in the way I have seen some of the answers written.

Re: lim x->infinity (1+(1/x))^x

By definition :

$\displaystyle e=\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x$

Re: lim x->infinity (1+(1/x))^x

Quote:

Originally Posted by

**princeps** By definition :

$\displaystyle e=\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x$

Thanks, although i think the whole question is about proving that that's true through using L' Hopital's Rule

Re: lim x->infinity (1+(1/x))^x

Quote:

Originally Posted by

**Cotty** Hi,

I have another question on limits:

lim (1+(1/x))^x

x->infinity

So i assume that I have to use L'Hopital's Rule to get to an answer

Firstly I have let y=the equation and taken the natural log of each side:

ln(y)=ln[(1/(1+(1/x)))^x]

and then made it the form:

=lim x*ln(1/(1+(1/x)))

x->infinity

=lim ln(1+(1+(1/x)))/x^-1

x->infinity

and used L'hopital to get to:

1/(1+(1/x))

Everywhere i have looked has said that the answer is that x approaches e and I have absolutely no idea how I get this solution? If you are able to give me a hand it would be greatly appreciated! Thanks, sorry it is so messy I don't know how to write it out in the way I have seen some of the answers written.

$\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty} \left[ \left( 1 + \frac{1}{x} \right)^x \right] &= \lim_{x \to \infty} e^{\ln{\left[ \left( 1 + \frac{1}{x} \right)^x \right]}} \\ &= \lim_{x \to \infty} e^{\frac{\ln{\left( 1 + \frac{1}{x} \right)}}{\frac{1}{x}}} \\ &= e^{\lim_{x \to \infty} \frac{\ln{\left( 1 + \frac{1}{x} \right)} }{\frac{1}{x}}} \end{align*}$

This limit goes to $\displaystyle \displaystyle \frac{0}{0}$ so now use L'Hospital's Rule.

Re: lim x->infinity (1+(1/x))^x

Quote:

Originally Posted by

**Prove It** $\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty} \left[ \left( 1 + \frac{1}{x} \right)^x \right] &= \lim_{x \to \infty} e^{\ln{\left[ \left( 1 + \frac{1}{x} \right)^x \right]}} \\ &= \lim_{x \to \infty} e^{\frac{\ln{\left( 1 + \frac{1}{x} \right)}}{\frac{1}{x}}} \\ &= e^{\lim_{x \to \infty} \frac{\ln{\left( 1 + \frac{1}{x} \right)} }{\frac{1}{x}}} \end{align*}$

This limit goes to $\displaystyle \displaystyle \frac{0}{0}$ so now use L'Hospital's Rule.

So is it correct that I used L'Hops to get x/(x+1) and again to get 1/1 and then put e^(1/1)?

Re: lim x->infinity (1+(1/x))^x

Quote:

Originally Posted by

**Cotty** So is it correct that I used L'Hops to get x/(x+1) and again to get 1/1 and then put e^(1/1)?

Yes that's correct.

Re: lim x->infinity (1+(1/x))^x

Thanks Prove It, you've been really helped me today!