# Math Help - limit

1. ## limit

evaluate $\lim_x\to$0 (e - (1+x) $^{1/x}$)/x
i used l'hopitals rule for the $\frac{o}{o}$ case but i get the limit as infinity but when i graph this function the lim is 1.25... any sort of insight would be helpful.

2. Originally Posted by bubble86
evaluate $\lim_x\to$0 (e - (1+x) $^{1/x}$)/x
i used l'hopitals rule for the $\frac{o}{o}$ case but i get the limit as infinity but when i graph this function the lim is 1.25... any sort of insight would be helpful.
$\displaystyle \lim_{x\to 0} \frac {e-(1+x)^{\frac {1}{x}} } {x} = \frac {e}{2}$

show what have you done

3. Originally Posted by bubble86
evaluate $\lim_x\to$0 (e - (1+x) $^{1/x}$)/x
i used l'hopitals rule for the $\frac{o}{o}$ case but i get the limit as infinity but when i graph this function the lim is 1.25... any sort of insight would be helpful.
To apply l'Hospital's Rule you need to be able to differentiate $(1 - x)^{1/x}$. I suggest writing it as $e^{\frac{1}{x} \ln (1 - x)$. Can you differentiate it now?

4. ## my work so far

$\lim_{x\to\o}$ $\ (e - ((1+x)^{\frac{1}{x}}$ ) / x
= $\lim_{x\to\o}$ $\frac{d}{dx}$ $\((1+x)^{\frac{1}{x}}$
= $\lim_{x\to\o}$ $\frac{d}{dx}$ $\ e^{\ln\((1+x)^{\frac{1}{x}}}$
= $\lim_{x\to\o}$ $\ e^{\ln\((1+x)^{\frac{1}{x}}}$ $\frac{d}{dx}$ $\ln\((1+x)^{\frac{1}{x}}$
= $\lim_{x\to\o}$ $\ e^{\ln\((1+x)^{\frac{1}{x}}}$ $\frac{d}{dx}$ $\frac{1}{x} \\ln\((1+x)$
= $\lim_{x\to\o}$ $\ e^{\ln\((1+x)^{\frac{1}{x}}}$ $\cdot(\frac{-1}{x^{2}}\ln(1+x)\ +\frac{1}{x}\cdot\frac{1}{1+x})$
= $\lim_{x\to\o} (1+x)^{\frac{1}{x}}\cdot\ ( ln (1+x)^{\frac{-1}{x^{2}}}\ + \frac{1}{x}\cdot\frac{1}{(1+x)})$
the term $\lim_{x\to\o}\frac{1}{x}}\cdot\frac{1}{1+x}$ tends to $\infty$ so does that mean the whole expression
$\lim_{x\to\o} (1+x)^{\frac{1}{x}}\cdot\ ( ln (1+x)^{\frac{-1}{x^{2}}}\ + \frac{1}{x}\cdot\frac{1}{(1+x)})$ tend to $\infty$

5. Originally Posted by bubble86
$\lim_{x\to\o}$ $\ (e - ((1+x)^{\frac{1}{x}}$ ) / x
= $\lim_{x\to\o}$ $\frac{d}{dx}$ $\((1+x)^{\frac{1}{x}}$
= $\lim_{x\to\o}$ $\frac{d}{dx}$ $\ e^{\ln\((1+x)^{\frac{1}{x}}}$
= $\lim_{x\to\o}$ $\ e^{\ln\((1+x)^{\frac{1}{x}}}$ $\frac{d}{dx}$ $\ln\((1+x)^{\frac{1}{x}}$
= $\lim_{x\to\o}$ $\ e^{\ln\((1+x)^{\frac{1}{x}}}$ $\frac{d}{dx}$ $\frac{1}{x} \\ln\((1+x)$
= $\lim_{x\to\o}$ $\ e^{\ln\((1+x)^{\frac{1}{x}}}$ $\cdot(\frac{-1}{x^{2}}\ln(1+x)\ +\frac{1}{x}\cdot\frac{1}{1+x})$
= $\lim_{x\to\o} (1+x)^{\frac{1}{x}}\cdot\ ( ln (1+x)^{\frac{-1}{x^{2}}}\ + \frac{1}{x}\cdot\frac{1}{(1+x)})$
the term $\lim_{x\to\o}\frac{1}{x}}\cdot\frac{1}{1+x}$ tends to $\infty$ so does that mean the whole expression
$\lim_{x\to\o} (1+x)^{\frac{1}{x}}\cdot\ ( ln (1+x)^{\frac{-1}{x^{2}}}\ + \frac{1}{x}\cdot\frac{1}{(1+x)})$ tend to $\infty$
The derivative is a product of two functions. The limit of the first function in the product is readily recognised as e. The limit of the second function is 1/2: see limit of 1&#47;&#40;x&#40;1&#43;x&#41;&#41; - Log&#91;1&#43;x&#93;&#47;x&#94;2 as x approaches 0 - Wolfram|Alpha (click on show steps).

6. The taylor series of ln(x+1) is
ln(1+x)=x-x^2/2+x^3/3-x^4/4+-...
for |x|<1
so
1/x^2 * ln(1+x)-1/x*1/(x+1)=
1/x^2*(x-x^2/2+x^3/3-x^4/4+-...)-1/x*1/(x+1)=
1/x-1/2+x*(...)-1/x*1/(x+1)=
1/x(1-1/(x+1))-1/2+x*(...)=
1/(x+1)-1/2+x*(...)->1/2 for x->0

7. Iondor I follow you until the 2nd last line. The taylor series for ln (1+x) is $\sum_{n=0}^{\infty}\ (-1)^n\frac{x^{n+1}}{n+1}$
so $\frac{-1}{x^2} ln (1+x) = \frac{-1}{x^2}\sum_{n=0}^{\infty}\ (-1)^n\frac{x^{n+1}}{n+1}$ this expands out to
$\frac{-1}{x^2}\cdot\ ( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} .....)$
$\ = (\frac{-1}{x} + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6})$
now $\lim_{x\to\o} (\frac{-1}{x} + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x})$
$\ = \lim_{x\to\o} (\frac{-1}{x}\cdot\frac{1+x}{1+x} + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x})$ is this what you mean?
in which case wouldn't the lim be $\ = \lim_{x\to\o} ( \frac{-1}{x}\cdot\frac{1}{1+x}\cdot (1+x) + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x})$
$\ = \lim_{x\to\o} (( \frac{-1}{x}\cdot\frac{1}{1+x} ) + (x\cdot\frac{-1}{x}\cdot\frac{1}{1+x} ) + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x})$
$\ = \lim_{x\to\o} ( (x\cdot\frac{-1}{x}\cdot\frac{1}{1+x} ) + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...) + (\frac{1}{x}\cdot\frac{1}{1+x}) + ( \frac{-1}{x}\cdot\frac{1}{1+x} )$
$\ = \lim_{x\to\o} ( ( \frac{-1}{1+x} ) + \frac{1}{2} - \frac{x}{3} +\frac{x^2}{4} - \frac{x^3}{5} + \frac{x^4}{6}...)$
$\ = -1 + \frac{1}{2}$
$\ = \frac{-1}{2}$

8. Originally Posted by bubble86
$\lim_{x\to\o}$ $\ (e - ((1+x)^{\frac{1}{x}}$ ) / x
= $\lim_{x\to\o}$ $\frac{d}{dx}$ $\((1+x)^{\frac{1}{x}}$
You missed here a minus sign, because you have to do differentiate the whole nominater, i.e. $
\frac{d}{dx} (e-(1+x)^{\frac{1}{x}})
=-\frac{d}{dx}(1+x)^{\frac{1}{x}}$

and from there you go as in your last post. You just have to multiply everything with -1 and then you get 1/2.

9. thanks a lot guys would have not taught the taylor series expansion on my own

10. $\displaystyle{\lim_{x\to 0} \frac{e-(1+x)^{\frac{1}{x}}}{x} = -\lim_{x\to 0} \frac{d}{dx} (1+x)^{\frac{1}{x}}= \lim_{x\to 0} (1+x)^{\frac{1}{x}} \frac{\ln(1+x)-x/(1+x)}{x^2}}$

$\displaystyle{=e \lim_{x\to 0} \frac{(1+x)^{-1}-(1+x)^{-2}}{2x}} = \frac{e}{2}$