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

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.
