Just added for your curiosity.
We can compute the antiderivative, starting with
$$\frac n {1+n^2x^2}=\frac n {(nx-i)(nx+i)}=\frac {ni} 2 \left(\frac 1 {nx+i}-\frac 1 {nx-i} \right)$$ which make that we face the problem of
$$I=\int \frac{\sin(x)}{x+a}\,dx$$ Let $x+a=t$ to make
$$I=\int \frac{\sin(t-a)}{t}\,dt=\cos(a)\int \frac{\sin(t)}{t}\,dt-\sin(a)\int \frac{\cos(t)}{t}\,dt $$ that is to say
$$I=\cos (a)\, \text{Si}(t)-\sin (a) \,\text{Ci}(t)$$ where appear the sine and cosine integrals (don' worry : sooner or later, you will leran about them).
Using the integration bounds and then the asymptotics, you will end with
$$I_n=\int_0^\infty \dfrac{n\sin x}{1+n^2x^2}\, dx=\frac{\log \left({n}\right)-\gamma +1}{n}+O\left(\frac{1}{n^3}\right)$$ which, for sure, shows the limit but also how it is approached. Moreover, as shown in the table below, this gives an approximation of the result
$$\left(
\begin{array}{ccc}
1 & 0.422784 & 0.646761 \\
2 & 0.557966 & 0.599204 \\
3 & 0.507132 & 0.521764 \\
4 & 0.452270 & 0.459176 \\
5 & 0.406444 & 0.410274 \\
6 & 0.369091 & 0.371446 \\
7 & 0.338385 & 0.339943 \\
8 & 0.312778 & 0.313865 \\
9 & 0.291112 & 0.291902 \\
10 & 0.272537 & 0.273130 \\
20 & 0.170926 & 0.171014 \\
30 & 0.127466 & 0.127495 \\
40 & 0.102792 & 0.102804 \\
50 & 0.086696 & 0.086703 \\
60 & 0.075286 & 0.075290 \\
70 & 0.066733 & 0.066735 \\
80 & 0.060060 & 0.060062 \\
90 & 0.054696 & 0.054697 \\
100 & 0.050280 & 0.050281
\end{array}
\right)$$