I want to compute the $\lim_{n\rightarrow \infty}E(e^{-\delta (t\wedge \tau_{n})}f(X_{t\wedge \tau_{n}}))$.
What I know is that $P(\tau_{n}\rightarrow \infty)=1$ and that $f(x)$ has quadratic growth i.e. $|f(x)|\leq C(1+|x|^{2})$. Also, $\lim_{n\rightarrow \infty}e^{-\delta (t\wedge \tau_{n})}f(X_{t\wedge \tau_{n}})=e^{-\delta t}f(X_{t})$ and $e^{-\delta t}X^{2}_{t}$ is integrable.
I apply the DCT the following way:
$$|e^{-\delta (t\wedge \tau_{n})}f(X_{t\wedge \tau_{n}})|\leq Ce^{-\delta (t\wedge \tau_{n})}(1+|X_{t\wedge \tau_{n}}|^{2})$$
For large enough $n$ the righthand side of the previous inequality is equal to $Ce^{-\delta t}(1+|X_{t}|^{2})$ which is integrable. Thus, applying the DCT we get
$$\lim_{n\rightarrow \infty}E(e^{-\delta (t\wedge \tau_{n})}f(X_{t\wedge \tau_{n}}))=E(e^{-\delta t}f(X_{t})$$
Is this correct? I know the DCT asks for the function to be bounded for all $n$. Is there another way to compute that limit?
Yes, that is correct. To elaborate, if $f_n \to f$ a.e. and there exists some $N \in \mathbb{N}$ such that $f_n$ is dominated by an integrable function $h$ for all $n > N$ then the conclusion of DCT still holds. To see this you can set $g_n = f_{n+N}$ and apply the usual version of DCT to $(g_n)_{n\in \mathbb{N}}$ to conclude (since $g_n \to f$) that $$\lim_{n\to \infty} E(g_n) = E(f)$$
But then $$E(f)=\lim_{n\to \infty}E(g_{n-N})=\lim_{n\to \infty}E(f_n)$$ proving the claim.