3

find the value of $$\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}$$I use hospital law and can't find answer

amir bahadory
  • 4,629
  • 1
    @Dr.MV It does now he put a minus sign in. – Frank Apr 23 '15 at 19:15
  • 1
    @Zach466920 How do you compute Taylor's expansion ? I will do it by differentiating the function, which is as ugly as using L'Hospital. – Alonso Delfín Apr 23 '15 at 19:27
  • Use the ones that are known: $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots+(-1)^{n+1}\frac{{x^n}}{n}+o(x^n)$, for instance. – Clement C. Apr 23 '15 at 19:29
  • 1
    @zach466920 Note that an edit was made after I posted. Please delete your comment. – Mark Viola Apr 23 '15 at 19:29
  • @AlonsoDelfín Clement C. made a better derivation than I did (I'd left it ambiguous), so I just deleted my answer. No need for L' Hopital's rule – Zach466920 Apr 23 '15 at 19:30

3 Answers3

6

Without using l'Hospital (I tend to avoid it as much as possible, as it always looked like a heavy hammer to me): $$ (1+x)^{1/x} = e^{\frac{1}{x}\ln(1+x)} = e^{\frac{1}{x}(x-\frac{x^2}{2}+o(x^2))} = e^{1-\frac{x}{2}+o(x))} = e(1+\frac{x}{2}+o(x)) = e-\frac{xe}{2}+o(x) $$ using the Taylor expansions of $\ln(1+x)$ and $e^x$ when $x\to0$. Plugging it back, the expression becomes $$ \frac{e-(e-\frac{xe}{2}+o(x))}{x} = \frac{\frac{xe}{2}+o(x)}{x} = \frac{e}{2}+o(1) $$ when $x\to0$. Hence, the limit is $\frac{e}{2}$.

Clement C.
  • 67,323
3

$$\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}\stackrel{l'H}=\lim_{x\to 0}\left(\frac1{x^2}\log(1+x)-\frac1{x(1+x)}\right)(1+x)^{1/x}=$$

$$=\lim_{x\to 0}\color{red}{(1+x)^{1/x}}\frac{(1+x)\log(1+x)-x}{x^2(1+x)}$$

Now, on the problematic factor above:

$$\lim_{x\to 0}\frac{(1+x)\log(1+x)-x}{x^2(1+x)}\stackrel{l'H}=\lim_{x\to0}\frac{\log(1+x)}{2x(1+x)+x^2}\stackrel{l'H}=$$

$$=\lim_{x\to0}\frac1{(1+x)\left(2(1+x)+4x\right)}=\frac1{1\cdot2}=\frac12$$

and together with the red factor above and arithmetic of limits, the limit is $\;\dfrac e2\;$ .

Timbuc
  • 34,191
2

A little different (again no L'Hopital): Set $g(x) = [\ln (1+x)]/x = 1-x/2+x^2/3 - \cdots .$ Our expression can be written

$$-\frac{e^{g(x)}-e^{g(0)}}{x-0} \to -(e^g)'(0) = -e^{g(0)}g'(0)= -e\cdot (-1/2) = e/2.$$

zhw.
  • 105,693