Let $f\geq 0$ continuous and integrable over $[0,\infty [$ and $g\geq 0$ continuous and bounded $\geq 0$ over $[0,\infty [$. Let $\lambda >0$. Set $$\varphi(\lambda )=\int_0^\infty \frac{f}{\lambda +g}.$$
Prove that $$\varphi(\lambda )\sim_{\lambda \to \infty }\frac{1}{\lambda }\int_0^\infty f.$$
Can I do as following : Since $g$ is bounded,
$$g(x)=\mathcal O(1),$$ and thus $$\lambda +g(x)=\lambda +\mathcal O(1)=\lambda \left(1+\mathcal O\left(\frac{1}{\lambda }\right)\right)=\lambda (1+o(1)).$$ Therefore $$\frac{f}{\lambda +g}=\frac{f}{\lambda \left(1+o(1)\right)}=\frac{f}{\lambda} \left(1+o(1)\right),$$ and thus $$\int_0^\infty \frac{f}{\lambda +g}=\frac{1}{\lambda} \int_0^\infty f+o\left(\frac{1}{\lambda }\int_0^\infty f\right),$$ and thus $$\varphi(\lambda )\sim_{\lambda \to \infty }\frac{1}{\lambda} \int_0^\infty f.$$
Is it working ?
Added
We have when $\lambda \to \infty $ that $$\frac{f}{\lambda+g}=\frac{f}{\lambda} + o\left(\frac{f}{\lambda }\right).$$ I set $R(\lambda,x )=o\left(\frac{f(x)}{\lambda }\right)$ Let $\varepsilon>0$. There is $M>0$ s.t. $$|R(\lambda,x )|\leq \varepsilon\frac{f}{\lambda },$$ when $\lambda \geq M$. Then $$\int_0^\infty |R(x,\lambda )|\leq \frac{\varepsilon}{\lambda }\int_0^\infty f(x)\,dx,$$ and thus $$\int_0^\infty R(x,\lambda )\,dx = o\left(\frac{1}{\lambda} \int_0^\infty f(x)\,dx\right).$$
There exists some $M$ such that, for every $x\geq 0$, $0\leq g(x)\leq M$.
For any $x$, $\displaystyle \left|\frac{1}{\lambda +g(x)}-\frac{1}{\lambda } \right|= \frac{g(x)}{\lambda(\lambda+g(x))}\leq \frac {g(x)}{\lambda^2}\leq \frac M{\lambda^2}$
Thus $$ \left|\int_0^{\infty} \frac{f}{\lambda +g}-\frac{f}{\lambda }\right|\leq \int_0^{\infty}f\left|\frac{1}{\lambda +g}-\frac{1}{\lambda } \right|\leq \frac{M}{\lambda^2}\int_0^{\infty}f$$
Hence $$\int_0^{\infty} \frac{f}{\lambda +g}=\int_0^{\infty}\frac{f}{\lambda } + O\left( \frac{1}{\lambda^2}\right)=\int_0^{\infty}\frac{f}{\lambda } + o\left( \frac{1}{\lambda}\right)$$