Given $f,g:[1,+\infty)\to\mathbb{R}$ s.t $\lim _{x\to+\infty}\frac{f(x)}{g(x)}=L>0$. Prove $\int_1^{+\infty}f,\int_1^{+\infty}g$ converge/diverge

Since we are able to talk about limit $L:=\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}$, there exists $X_{1}>0$ such that $g(x)\neq0$ whenever $x\geq X_{1}$. Note that $g$ is continuous, by Intermediate Value Theorem, we further have $g(x)>0$ for all $x\geq X_{1}$ or $g(x)<0$ for all $x\geq X_{1}$. Without loss of generality, we may assume that $g(x)>0$ for all $x\geq X_{1}$ (otherwise replace $f$ and $g$ by $-f$ and $-g$ respectively). Since $0<\frac{L}{2}<\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}<2L$, we can choose $X_{2}>0$ such that $\frac{L}{2}<\frac{f(x)}{g(x)}<2L$ whenever $x\geq X_{2}$. Take $X=\max(X_{1},X_{2})$. Note that $f(x)>0$ and $g(x)>0$ whenever $x\geq X$.

Define $F(x)=\int_{X}^{x}f(t)dt$ and $G(x)=\int_{X}^{x}g(t)dt$. Note that $F$ and $G$ are monotonic increasing. Observe that $F(x)\leq2L\cdot G(x)$ and $G(x)\leq\frac{2}{L}F(x)$, so either $\lim_{x\rightarrow\infty}F(x)<\infty$ and $\lim_{x\rightarrow\infty}G(x)<\infty$ or $\lim_{x\rightarrow\infty}F(x)=\infty$ and $\lim_{x\rightarrow\infty}G(x)=\infty$. That is, either both improper integrals converge or both improper integrals diverge.