# limit

• Mar 8th 2008, 05:22 PM
akhayoon
limit
$\lim{x}\rightarrow{\infty} \frac{(1+x)^{\frac{1}{x}}-e}{x}$

well the answer is -e/2, but I can't seem to get the answer...mostly due the fact that there is a 1/x power on the 1+x polynomial and how to implement it with
L'Hopitale's rule...?
• Mar 8th 2008, 05:50 PM
mr fantastic
Quote:

Originally Posted by akhayoon
$\lim{x}\rightarrow{\infty} \frac{(1+x)^{\frac{1}{x}}-e}{x}$

well the answer is -e/2, but I can't seem to get the answer...mostly due the fact that there is a 1/x power on the 1+x polynomial and how to implement it with
L'Hopitale's rule...?

Well, $\lim_{x \rightarrow \infty} (1 + x)^{1/x} = e$ so you can use l'Hopital's rule.

To get the derivative of $y = (1 + x)^{1/x}$ you can use logarithmic differentiation.

Edit: My mistake: $\lim_{x \rightarrow \infty} (1 + x)^{1/x} = 1$, NOT e. l'H CANNOT be used. See post #4.
• Mar 8th 2008, 05:59 PM
akhayoon
still kind of shaky on this I tried multiplying everything by ln so I could reduce the
$(1+x)^{\frac{1}{x}}$ with the ln rule, but everything gets kind of messy and I end up getting a function that could never give me $e$
• Mar 8th 2008, 06:08 PM
mr fantastic
Quote:

Originally Posted by akhayoon
still kind of shaky on this I tried multiplying everything by ln so I could reduce the
$(1+x)^{\frac{1}{x}}$ with the ln rule, but everything gets kind of messy and I end up getting a function that could never give me $e$

I made a careless mistake in the limit and have edited my reply. As posted, the value of the limit is zero. Have you posted the correct expression? Perhaps x --> 0, not oo?
• Mar 8th 2008, 06:54 PM
mr fantastic
Quote:

Originally Posted by akhayoon
still kind of shaky on this I tried multiplying everything by ln so I could reduce the
$(1+x)^{\frac{1}{x}}$ with the ln rule, but everything gets kind of messy and I end up getting a function that could never give me $e$

The limit >is< meant to be x --> 0. $\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}} = e$ so l'Hospital's Rule can be used.

To get the derivative of $y = (1 + x)^{1/x}$:

$\ln y = \frac{1}{x} \ln (1 + x)$

$\Rightarrow \frac{1}{y}\, \frac{dy}{dx} = - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)}$

$\Rightarrow \frac{dy}{dx} = y \left( - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)} \right)$.

So the limit becomes:

$\lim_{x \rightarrow 0} (1 + x)^{1/x} \left( - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)} \right) = e \lim_{x \rightarrow 0} \left( - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)} \right)$.

Now show that $\lim_{x \rightarrow 0} \left( - \frac{1}{x^2} \ln (1 + x) + \frac{1}{x(1 + x)} \right) = \lim_{x \rightarrow 0} \left(\frac{x - (1 + x) \, \ln (1 + x)}{x^2(1 + x)} \right) = - \frac{1}{2}$.

Multiple applications of l'Hospital's rule is one possible approach.
Another approach is to substitute the series expansion for ln (1 + x): $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ......$. This substitution is valid since the series converges for $-1 < x \leq 1$ and $x \rightarrow 0$ ......
• Mar 9th 2008, 06:56 AM
akhayoon
ok, so I evaluate the limit as it approaches zero to get -e/2 like u showed, but aren't I asked to evaluate it as it approaches infinity? so how does this cover that?
• Mar 9th 2008, 01:39 PM
mr fantastic
Quote:

Originally Posted by akhayoon
ok, so I evaluate the limit as it approaches zero to get -e/2 like u showed, but aren't I asked to evaluate it as it approaches infinity? so how does this cover that?

I have already addressed that question - read post #4 again: In particular the bit where I say that if x --> oo the limit is equal to 0.

Clearly the limit is meant to be x --> 0. Clearly there is a typographical error in the question.