Jtdylktuy

Evaluate $limlimits_{x to 0} frac{(1+x)^{1/x} - e + frac{1}{2}ex}{x^2}$

One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{log_e (1+x)^{1/x}} = e^{frac{1}{x} (x-frac{x^2}{2} -frac{x^3}{3}+...)}$$
and then after complete expansion of each and every and substuting into back to limit and solving I get $frac{11e}{24}$ as an answer.

Now, this is a relatively long and complicated way to solve as you can see. I want to know if there is an easier way to solve this problem. Please help. Thank you!

Write $f(x) = frac{1}{x}log(1+x)$ and $f(0) = 1$. We know that $f$ so defined is analytics near $0$. Now, by the L'Hospital's rule applied twice,

begin{align*}
lim_{xto0} frac{e^{f(x)} - e + frac{e}{2}x}{x^2}
&= lim_{xto0} frac{e^{f(x)}f'(x) + frac{e}{2}}{2x} \
&= lim_{xto0} frac{e^{f(x)}f''(x) + e^{f(x)}f'(x)^2}{2} \
&= frac{e}{2}f''(0) + frac{e}{2}f'(0)^2.
end{align*}

Since $ f(x) = 1 - frac{1}{2}x + frac{1}{3}x^2 + cdots $ near $0$, it follows that $f'(0) = -frac{1}{2}$ and $f''(0) = frac{2}{3}$. Therefore the limit equals

$$ frac{e}{2}cdotfrac{2}{3} + frac{e}{2}left(-frac{1}{2}right)^2
= frac{11}{24}e. $$

If I may suggest, the problem of
$$y=frac{(1+x)^{frac1 x} - e + frac{1}{2}ex}{x^2}$$ is not so difficult if you use another way.
$$a=(1+x)^{frac1 x}implies log(a)= {frac1 x}log(1+x)$$
$$ log(a)={frac1 x}left(x-frac{x^2}{2}+frac{x^3}{3}-frac{x^4}{4}+Oleft(x^5right) right)=1-frac{x}{2}+frac{x^2}{3}-frac{x^3}{4}+Oleft(x^4right)$$ Now, continuing with Taylor
$$a=e^{log(a)}=e-frac{e x}{2}+frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right)$$
$$y=frac{frac{11 e x^2}{24}-frac{7 e x^3}{16}+Oleft(x^4right) }{x^2}=frac{11 e}{24}-frac{7 e x}{16}+Oleft(x^2right)$$ which gives not only the limit but also how it is approached.

Solution without expansions by the L'Hospital's rule only:
$$lim_{xrightarrow0}frac{(1+x)^{1/x} - e + frac{1}{2}ex}{x^2}=lim_{xrightarrow0}frac{(1+x)^{frac{1}{x}}left(frac{ln(1+x)}{x}right)'+frac{1}{2}e}{2x}=$$
$$=lim_{xrightarrow0}frac{frac{(1+x)^{frac{1}{x}}left(frac{x}{1+x}-ln(1+x)right)}{x^2}+frac{1}{2}e}{2x}=lim_{xrightarrow0}frac{left(frac{(1+x)^{frac{1}{x}}left(frac{x}{1+x}-ln(1+x)right)}{x^2}right)'}{2}$$ because
$$lim_{xrightarrow0}frac{(1+x)^{frac{1}{x}}left(frac{x}{1+x}-ln(1+x)right)}{x^2}=elim_{xrightarrow0}frac{x-(1+x)ln(1+x)}{x^2+x^3}=$$
$$=elim_{xrightarrow0}frac{1-ln(1+x)-1}{2x+3x^2}=-frac{e}{2}$$ and we can continue:
$$ lim_{xrightarrow0}frac{left(frac{(1+x)^{frac{1}{x}}left(frac{x}{1+x}-ln(1+x)right)}{x^2}right)'}{2}=$$
$$=frac{1}{2}lim_{xrightarrow0}left(frac{frac{(1+x)^{frac{1}{x}}left(frac{x}{1+x}-ln(1+x)right)^2}{x^2}+(1+x)^{frac{1}{x}}left(frac{1}{(1+x)^2}-frac{1}{1+x}right)}{x^2}-frac{2}{x^3}(1+x)^{frac{1}{x}}left(frac{x}{1+x}-ln(1+x)right)right)=$$
$$=frac{e}{2}lim_{xrightarrow}left(frac{frac{left(frac{x}{1+x}-ln(1+x)right)^2}{x^2}-frac{x}{(1+x)^2}}{x^2}-frac{2}{x^3}left(frac{x}{1+x}-ln(1+x)right)right)=$$
$$=frac{e}{2}lim_{xrightarrow0}left(-frac{3x+1}{x^2(1+x)^2}+frac{2ln(1+x)}{(1+x)x^2}+frac{ln^2(1+x)}{x^4}right)=$$
$$=frac{e}{2}lim_{xrightarrow0}frac{(1+x)^2ln^2(1+x)+(2x^3+2x^2)ln(1+x)-3x^3-x^2}{x^4}=...=frac{11e}{24}.$$