Let's being by rewriting the binomial distribution in terms of $\lambda$
$$
f_X(x) = \binom{n}{x} \left( \frac{\lambda}{n} \right)^x \left( 1 -
\frac{\lambda}{n} \right)^{n-x} = \frac{n!}{x!(n-x)!} \left( \frac{\lambda}{n}
\right)^x \left( 1 -
\frac{\lambda}{n} \right)^{n-x}
$$
And now lets use the limit from the question:
$$
\begin{aligned}
\lim_{n \rightarrow \infty} \frac{n!}{x!(n-x)!} \left( \frac{\lambda}{n}
\right)^x \left( 1 - \frac{\lambda}{n} \right)^{n-x} & = \lim_{n \rightarrow
\infty} \frac{\lambda^x n^x}{n^x x!} \left( 1- \frac{\lambda}{n} \right)^{n-x}
\\
& = \frac{\lambda^x}{x!} \lim_{n\rightarrow \infty} \left( 1- \frac{\lambda}{n}
\right)^n \\
& = \frac{\lambda^x}{x!} e^{-\lambda}
\end{aligned}
$$
In the last line here we need to use the Euler limit definition of exponential function
$e^{-\lambda} = \lim_{n\rightarrow \infty} \left( 1- \frac{\lambda}{n} \right)^n$
Notice that in answering this question we have had to use a result from the algebra
of limits. In particular, we have had to use the fact that if $\lim_{x\rightarrow a} f(x) = L_f$
and $\lim_{x\rightarrow a} g(x) = L_g$ then:
$$
\lim_{x\rightarrow a} f(x)g(x) = \lim_{x\rightarrow a} f(x)\lim_{x\rightarrow a} g(x) = L_f L_g
$$
