Consider some sequences $\{a_n\}, \{b_n\}, \{c_n\}$. Suppose $$ \begin{cases} \lim_{n\rightarrow \infty} (a_n+b_n)=0\\ \lim_{n\rightarrow \infty}c_n=d \end{cases} $$ for some $d\in \mathbb{R}$.
I want to show that $$ \exp(a_n)c_n=\exp(-b_n)d+o(1) $$
What I want to show is equivalent to showing that $$ \lim_{n\rightarrow \infty} [\exp(a_n)c_n-\exp(-b_n)d]=0 $$ I don't know how to proceed from here. Any hint would be extremely appreciated.
On
This is not even true for constant $c_n = 1$. With $a_n = n$, $b_n = \frac 1n - n$ we have $$ \exp(a_n) - \exp(-b_n) = \exp(n) - \exp(n - \frac 1n) = \exp(n) \bigl(1 - \exp(-\frac 1n) \bigr) \to \infty $$ since $1 - \exp(-\frac 1n) \sim \frac 1n$ for $n \to \infty$.
However, it becomes true if $(a_n)$ – or equivalently, $(b_n)$ – is assumed to be a bounded sequence: $$ \exp(a_n) c_n - \exp(-b_n) d = \bigl(1 - \exp(-a_n - b_n) \bigr) \exp(a_n) c_n + (c_n - d) \exp(-b_n) \\ \to 0 $$ because $$1 - \exp(-a_n - b_n) \to 0 \\ c_n - d \to 0$$ and all other factors are bounded.
As @MartinR points out, the statement is not true for $a_n, b_n, c_n \in \mathbb R$. A counterexample is $c_n=\frac 1n$, $a_n = n$, $b_n = -n$.
The statement is true, however, if we assume $a_n, b_n \ge 0$. In which case we know that both $$\lim_{n \to \infty} a_n = 0$$ $$\lim_{n \to \infty} b_n = 0$$
Therefore the last limit is easy to compute, as all limits exists:
$$\lim_{n\rightarrow \infty} [\exp(a_n)c_n-\exp(-b_n)d]= $$ $$\lim_{n\rightarrow \infty} \exp(a_n) \cdot \lim_{n\rightarrow \infty} c_n - d\lim_{n\rightarrow \infty} \exp(-b_n) = 1 \cdot d - d \cdot 1 = 0$$