Evaluate $\lim_{n \rightarrow \infty} \int_{-n}^n f(1+\frac{x}{n^2}) g(x) dx$

I want to evaluate $\lim_{n \rightarrow \infty} \int_{-n}^n f(1+\frac{x}{n^2}) g(x) dx$, where $g: \mathbb{R} \rightarrow \mathbb{R}$ is (Lebesgue)-integrable, and $f:\mathbb{R} \rightarrow \mathbb{R}$ is bounded, measurable and continuous at 1.

Approach 1 (confident about this)

My first approach is to use DCT, as follows.

1. First, for all $x \in \mathbb{R}$, $f(1+\frac{x}{n^2}) 1_{(-n,n)} \rightarrow f(1)$. Thus, the limit is ideally $\int_{\mathbb{R}} f(1) g(x) dx$.
2. Since $f$ is bounded, $|f(1+\frac{x}{n^2}) g(x)|$< $M|g(x)|$, where the RHS is integrable, so that by DCT, the above limit is$\int_{\mathbb{R}} f(1) g(x) dx$.

Approach 2 (not so confident about this)

My second approach was to note the $n^2$ denominator term and try out more "basic" analysis, since the above approach also seems to work with $n$ in the denominator term? However, I am doubtful about some steps, which I've typed in boldface. Please let me know if those steps are correct or not

1. For $x \in [-n,n]$, we have that $\frac{x}{n^2} \rightarrow 0$ uniformly in $x$. Since $f$ is continuous at $1$, this yields that $f(1+ \frac{x}{n^2}) \rightarrow f(1)$ uniformly on $[-n,n]$. Does this imply that $f(1+ \frac{x}{n^2}) 1_{(-n,n)}(x) \rightarrow f(1)$ uniformly in $x$, too?(I don't think it does, since $x \notin (-n,n) \implies f(1+ \frac{x}{n^2}) 1_{(-n,n)}(x) = 0$)And if it doesn't, then what that means is that "uniformly on $[-n,n]$" is not the same as simply attaching an indicator on that interval...

Supposing, however, that it is the same as attaching an indicator: Since we have a uniform limit now, it'd be nice to express the integral as a limit and then commute things. Hence, since g is integrable, so that $\int g = \int g_n$ for simple functions $g_n$, we have that (working backwards):

$$ \int f(1) g = \int \lim_{n \rightarrow \infty} f(1 + x/n^2) 1_{(-n,n)} g(x) dx = \lim_{m \rightarrow \infty} \int \lim_{n \rightarrow \infty} f(1 + x/n^2) 1_{(-n,n)} g_m(x) dx $$

$$ = \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \int f(1 + x/n^2) 1_{(-n,n)} g_m(x) dx = \lim_{n \rightarrow \infty} \int f(1 + x/n^2) 1_{(-n,n)} g(x) dx $$

, where we can commute the limits since (i) the limit with respect to $n$ is uniformly convergent, and (ii) the integral following the second equality sign is really a finite sum.