Limits at its maximum

Let $\mathfrak{L}$ denote the given limit. Doing some manupulations we get:- $\mathfrak{L}=\displaystyle\lim _{ n\rightarrow -1 }\dfrac{n+n^6+(n+1)\color{#69047E}{\frac{\ln(1+n+1)}{(n+1)}}}{\color{#D61F06}{\left(\frac{e^{\frac{n+1}{4}}-1}{\frac{n+1}{4}}\right)}\left(\frac{n+1}{4}\right)+1+\color{#3D99F6}{\frac{\ln(1+4n+4)}{4n+4}}(4n+4) +n^3 }$ $\boxed{\text{Using:-}\\\color{#0C6AC7}{ \displaystyle\lim_{x\to 0}\dfrac{e^x-1}{x}=1,\displaystyle\lim_{x\to 0}\dfrac{\ln(1+x)}{x}=1}}$ $\implies \mathfrak L=\displaystyle\lim_{n\to -1} ~4\left(\dfrac{n^6+2n+1}{4n^3+17n+21}\right)$ $\implies \mathfrak L=\displaystyle\lim_{n\to -1} 4\left(\dfrac{\xcancel{(n+1)}(n^5-n^4+n^3-n^2+n+1}{\xcancel{(n+1)}(4n^2-4n+21)}\right)$ $\mathfrak L=-\dfrac{16}{29}$ $\therefore\Huge29+16=\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{45}}}}}$

Let the limit be $\lambda$ . We note that the nominator and denominator functions both tend to $0$ as $x \to -1$ so we can use the L'Hôpital's rule.

$\begin{aligned} \lambda & = \lim_{x \to -1} \frac{x+x^6+\ln(x+2)}{e^{\frac{x+1}{4}}+\ln(4x+5)+x^3} \\ & = \lim_{x \to -1} \frac{1+6x^5+\frac{1}{x+2}}{\frac{e^{\frac{x+1}{4}}}{4}+\frac{4}{4x+5}+3x^2} \\ & = \frac{1-6+1}{\frac{1}{4}+4+3} = - \frac{16}{29} \end{aligned}$

$\Rightarrow A+B = 16+29 = \boxed{45}$