Evaluation of $\lim\limits_{x\rightarrow 0}\frac1x\left((1+2x+3x^2)^{1/x}-(1+2x-3x^2)^{1/x}\right) $ 
Evaluation of $$\lim_{x\rightarrow 0}\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} $$

$\bf{My\; Try::}$ Let $$l=\lim_{x\rightarrow 0}\frac{e^{\frac{\ln(1+2x+3x^2)}{x}}-e^{\frac{\ln(1+2x-3x^2)}{x}}}{x}$$
Using $$\bullet \; \frac{\ln(1+x)}{x}=x-\frac{x^2}{2}+\frac{x^3}{3}-.....\infty$$
But I am not Getting answer.
Now How can I solve after that, Help me
Thanks
 A: As usual this limit can also be evaluated without the use of L'Hospital's Rule and Taylor's series just by applying standard limits combined with the use of algebra of limits. We have
\begin{align}
L &= \lim_{x \to 0}\frac{(1 + 2x + 3x^{2})^{1/x} - (1 + 2x - 3x^{2})^{1/x}}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + 2x + 3x^{2})}{x}\right) - \exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{x}\right)}{x}\notag\\
&= \lim_{x \to 0}\exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{x}\right)\cdot\dfrac{\exp\left(\dfrac{\log(1 + 2x + 3x^{2}) - \log(1 + 2x - 3x^{2})}{x}\right) - 1}{x}\notag\\
&= \lim_{x \to 0}\exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{2x - 3x^{2}}\cdot(2 - 3x)\right)\cdot\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right)\right) - 1}{x}\notag\\
&= \exp(1\cdot 2)\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right)\right) - 1}{x}\notag\\
&= \exp(2)\lim_{z \to 0}\frac{\exp(z) - 1}{z}\cdot\lim_{x \to 0}\dfrac{z}{x}\notag\\
&= \exp(2)\lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{6x^{2}}{1 + 2x - 3x^{2}}\right)}{x^{2}}\notag\\
&= \exp(2)\lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{6x^{2}}{1 + 2x - 3x^{2}}\right)}{\dfrac{6x^{2}}{1 + 2x - 3x^{2}}}\cdot\frac{6}{1 + 2x - 3x^{2}}\notag\\
&= e^{2}\cdot 1\cdot 6 = 6e^{2}\notag
\end{align}
We have used the standard limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1 = \lim_{x \to 0}\frac{e^{x} - 1}{x}$$ and the fact that $$z = \frac{1}{x}\log\left(\frac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right) = \frac{1}{x}\log\left(1 + \frac{6x^{2}}{1 + 2x - 3x^{2}}\right)\to 0$$ as $x \to 0$ (which easily follows from the standard limits mentioned above).
A: Consider $$A=\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} =\frac{B^{\frac{1}{x}}-C^{\frac{1}{x}}}x$$ using $$B=(1+2x+3x^2) \qquad , \qquad C=(1+2x-3x^2)$$ So $$\log(B^{\frac{1}{x}})=\frac{1}{x}\log(B)$$ and now, using Taylor series $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ replace $y$ by $(2x+3x^2)$ to get $$\log(B)=2 x+x^2-\frac{10 x^3}{3}+O\left(x^4\right)$$ which gives $$\log(B^{\frac{1}{x}})=2+x-\frac{10 x^2}{3}+O\left(x^3\right)$$ Now, using Taylor again, $$B^{\frac{1}{x}}=e^{\log(B^{\frac{1}{x}})}=e^2+e^2 x-\frac{17 e^2 x^2}{6}+O\left(x^3\right)$$ Doing the same with the second term, you should arrive to $$C^{\frac{1}{x}}=e^2-5 e^2 x+\frac{127 e^2 x^2}{6}+O\left(x^3\right)$$ All of that makes $$A=6 e^2-24 e^2 x+O\left(x^2\right)$$ which shows the limit and how it is approached.
A: Hint: $e^{1+t} - e^{1-t} \sim 2t$ as t tends to 0
