sequences and series - $lim_{nrightarrow infty} (a_n+b_n)=0$ and $lim_{nrightarrow infty} c_n=L$ imply $lim_{nrightarrow infty} exp(a_n)*c_n-L*exp(-b_n)=0$

Consider the sequences $\{a_n\}_{\forall n \in \mathbb{N}}<0$, $\{b_n\}_{\forall n \in \mathbb{N}}>0$, $\{c_n\}_{\forall n \in \mathbb{N}}>0$ and suppose
$$
\begin{cases}
\lim_{n\rightarrow \infty} (a_n+b_n)=0\\
\lim_{n\rightarrow \infty} c_n=L<\infty
\end{cases}
$$

Could you help me to show that
$$

\lim_{n\rightarrow \infty} [\exp(a_n)*c_n-L*\exp(-b_n)]=0
$$
?

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I know that by assumption
$$
\lim_{n\rightarrow \infty} [\exp(a_n)*c_n-L*\exp(-b_n)]= \lim_{n\rightarrow \infty} [\exp(-b_n+o(1))*(L+o(1))-L*\exp(-b_n)]
$$

where $o(1)$ is a number going to zero as $n\rightarrow \infty$. How can I proceed from here?

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Let me add another assumption (thanks to a comment below)
$$
\exp(a_n)\equiv \Pi_{k=1}^{2n} x_{n,k}
$$
where $x_{n,k}\in [0,1]$ and $\lim_{n\rightarrow \infty} x_{n,k}=1$ $\forall k$

Answer

You can just compute
$$ |e^{a_n}c_n - L e^{-b_n}| = e^{a_n}|c_n - L e^{-(a_n+b_n)}| \leq |c_n - L e^{-(a_n+b_n)}|. $$
Now $c_n\to L$ and $a_n+b_n\to 0$. So, because $e^x$ and $|x|$ are continuous functions, you can "pass to the limit inside the functions" and get
$$ |e^{a_n}c_n - L e^{-b_n}| \leq |c_n - L e^{-(a_n+b_n)}| \to |L-Le^{-0}| = 0. $$
By the comparison principle, you get convergence of that guy to 0.