solving, the following limit So, the following is the question  given:

I could solve it partially, here is my approach:
The limit is of the form $(A+B)/C$ where $A$ and $B$ both approach $e^3$ while $C$ approaches $0$,
This can be found out by simply evaluating $A$ and $B$ separately.
Now, we can write the limit as
$$ \lim_{t \to 0}  [(1+3t+2t^2)^{1/t} - e^3]/t -\lim_{t \to 0} [(1+3t-2t^2)^{1/t} - e^3]/t $$
but I couldn't evaluate these two limits at least using LH rule as the derivative of numerator is quite a long expression. Kindly suggest a way to solve this question, all help is greatly appreciated.
 A: \begin{align}
&\lim_{t\to0}\frac{1}{t}\left[(1+3t+2t^2)^{1/t}-(1+3t+2t^2)^{1/t}\right]=\\
&\qquad=\lim_{t\to0}\frac{1}{t}\left[(1+3t+2t^2)^{\frac{1}{3t+2t^2}\frac{3t+2t^2}{t}}-(1+3t-2t^2)^{\frac{1}{3t-2t^2}\frac{3t-2t^2}{t}}\right]=\\
&\qquad=\lim_{t\to0}\frac{1}{t}\left[e^{\frac{3t+2t^2}{t}}-e^{\frac{3t-2t^2}{t}}\right]=\\
&\qquad=e^3\lim_{t\to0}\frac{1}{t}\left[e^{2t}-e^{-2t}\right]=\\
&\qquad=2e^3\lim_{t\to0}\left[\frac{e^{2t}-1}{2t}+\frac{e^{-2t}-1}{-2t}\right]=4e^3
\end{align}
A: $$A=(1+3t+2t^2)^{\frac 1 t}\implies \log(A)=\frac 1 t \log(1+3t+2t^2)$$
$$ \log(1+3t+2t^2)=3 t-\frac{5 t^2}{2}+3 t^3-\frac{17 t^4}{4}+O\left(t^5\right)$$
$$ \log(A)=3-\frac{5 t}{2}+3 t^2-\frac{17 t^3}{4}+O\left(t^4\right)$$
$$A=e^{\log(A)}=e^3\left(1-\frac{5 t}{2}+\frac{49 t^2}{8}-\frac{689 t^3}{48}\right)+O\left(t^4\right) $$
$$B=(1+3t-2t^2)^{\frac 1 t}\implies \log(B)=\frac 1 t \log(1+3t-2t^2)$$
$$ \log(1+3t-2t^2)=3 t-\frac{13 t^2}{2}+15 t^3-\frac{161 t^4}{4}+O\left(t^5\right)$$
$$ \log(B)=3-\frac{13 t}{2}+15 t^2-\frac{161 t^3}{4}+O\left(t^4\right)$$
$$B=e^{\log(B)}=e^3\left(1-\frac{13 t}{2}+\frac{289 t^2}{8}-\frac{8809 t^3}{48} \right)+O\left(t^4\right) $$
$$A-B=4 e^3 t-30 e^3 t^2+\frac{1015 e^3 t^3}{6}+O\left(t^4\right)$$
$$\frac{A-B}t=4 e^3 -30 e^3 t+\frac{1015 e^3 t^2}{6}+O\left(t^3\right)$$ shows the limit and how it is approached.
A: Using Taylor's Theorem on the natural logarithm and exponential we have that
\begin{align}
(1+3t+2t^2)^{1/t}
&=\exp{\left(\frac{\ln{(1+3t+2t^2)}}t\right)}\\
&=\exp{\left(\frac{(3t+2t^2)-(3t+2t^2)^2/2+o(t^2)}t\right)}\\
&=\exp{\left(3-\frac52t+o(t)\right)}\\
&=e^3\exp{\left(-\frac52t+o(t)\right)}\\
&=e^3\left(1-\frac52t+o(t)\right)
\end{align}
and similarly we have
$$(1+3t-2t^2)^{1/t}=e^3\left(1-\frac{13}2t+o(t)\right)$$
So your limit is just
\begin{align}
\lim_{t\to0}\frac{e^3\left(1-\frac52t+o(t)\right)-e^3\left(1-\frac{13}2t+o(t)\right)}t
&=\lim_{t\to0}\frac{4e^3t+o(t)}t\\
&=\lim_{t\to0}(4e^3+o(1))\\
&=\boxed{4e^3}\\
\end{align}
