= Solution
The independent identically distributed <central limit theorem> states that for variables of common finite mean $\mu$ and finite positive <variance> $\sigma^2$,
$$
\frac{\sum_{i=1}^nX_i-n\mu}{\sigma\sqrt n}\ \xrightarrow{d}\ N(0,1).
$$
For the <exponential distribution>, $\mu=1/\lambda$ and $\sigma=1/\lambda$. Thus the event in question is exactly
$$
\left\{\frac{\lambda\sum_{i=1}^nX_i-n}{\sqrt n}<x\right\}.
$$
The limiting <normal distribution> has a continuous distribution function at every real $x$, so <convergence in distribution> gives
$$
\boxed{\mathbb P\left(\sum_{i=1}^nX_i<\frac{n+x\sqrt n}{\lambda}\right)\longrightarrow\Phi(x)}.
$$
Here $\Phi(x)=(2\pi)^{-1/2}\int_{-\infty}^xe^{-u^2/2}\,du$, including the strict-inequality convention because the limiting distribution is continuous.
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