Evaluation of $\lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}+(1+2x)^{\frac{1}{2x}}+(1+3x)^{\frac{1}{3x}}-3e}{9x}$ 
Evaluation of $\displaystyle \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}+(1+2x)^{\frac{1}{2x}}+(1+3x)^{\frac{1}{3x}}-3e}{9x}$
Without Using Series expansion and L Hopital Rule.

I have solve it Using Series expansion Like $\displaystyle \lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}} = \lim_{x\rightarrow 0}e^{\frac{\ln(1+x)}{x}}=\lim_{x\rightarrow 0}e^{\frac{x-\frac{x^2}{2}+\frac{x^3}{3}}{x}}$
So we get $\displaystyle \lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e^{1-\frac{x}{2}}$ Similarly for other terms.
But I did not understand How can we solve it Without Using Series expansion and L Hopital Rule.
Help Required, Thanks
 A: Let's calculate the limit $$f(k) = \lim_{x \to 0}\frac{(1 + kx)^{1/kx} - e}{x}\tag{1}$$ for positive $k$. The answer to your question is clearly $$L = \frac{f(1) + f(2) + f(3)}{9}\tag{2}$$ We have
\begin{align}
f(k) &= \lim_{x \to 0}\frac{(1 + kx)^{1/kx} - e}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + kx)}{kx}\right) - \exp(1)}{x}\notag\\
&= \exp(1)\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + kx)}{kx} - 1\right) - 1}{x}\notag\\
&= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + kx)}{kx} - 1\right) - 1}{\dfrac{\log(1 + kx)}{kx} - 1}\cdot\dfrac{\dfrac{\log(1 + kx)}{kx} - 1}{x}\notag\\
&= \frac{e}{k}\lim_{x \to 0}\frac{\log(1 + kx) - kx}{x^{2}}\notag\\
&= ke\lim_{x \to 0}\frac{\log(1 + kx) - kx}{k^{2}x^{2}}\notag\\
&= ke\lim_{t \to 0}\frac{\log(1 + t) - t}{t^{2}}\text{ (putting }t = kx)\notag\\
&= -\frac{ke}{2}\notag
\end{align}
The last limit can be evaluated without Taylor series or L'Hospital Rule but with some more effort as shown in this answer.
Then from $(2)$ we get desired limit $L = -e/3$.
A: You do have in general, because of $$\lim_{x\to 0}\left(\frac{(1+kx)^{\frac{1}{kx}}-e}{x}\right)= \frac{-ke}{2}$$ that
 $$\sum_{k=1}^n\lim_{x\to 0}\left(\frac{(1+kx)^{\frac{1}{kx}}-e}{x}\right)=\sum_{k=1}^n\frac{-ke}{2}=\frac{-n(n+1)e}{4}$$
However I can't to prove this without series or L'Hôpital Rule, it seems impossible to me.
You can easily verify  for the particular case proposed by the OP that the corresponding answer for $n=3$ is $$\frac{-3(4)e}{4\cdot9}=\color{red}{\frac{-e}{3}}$$
