If $\lim_{x\to a}f(x)=\infty$ and $\lim_{x\to a}g(x)=\infty$ then $\lim_{x\to a}\frac{f(x)}{g(x)}=1\iff\lim_{x\to a}(f(x)-g(x))=0$

Question

I want to prove the following statement

If $\lim_{x\to a}f(x)=\infty$ and $\lim_{x\to a}g(x)=\infty$ then $\lim_{x\to a}\frac{f(x)}{g(x)}=1\iff\lim_{x\to a}(f(x)-g(x))=0$

One side of the proof is easy. Suppose that $\lim_{x\to a}(f(x)-g(x))=0$. By our assumption we also know that $\lim_{x\to a}\frac{1}{g(x)}=0$ then we have

$$0=\lim_{x\to a}(f(x)-g(x))\lim_{x\to a}\frac{1}{g(x)}=\lim_{x\to a}\frac{f(x)-g(x)}{g(x)}=\lim_{x\to a}\frac{f(x)}{g(x)}-1$$ which is equivalent to $$\lim_{x\to a}\frac{f(x)}{g(x)}=1.$$ Now, let us go the other side. Suppose that $\lim_{x\to a}\frac{f(x)}{g(x)}=1$. This is again equivalent to $\lim_{x\to a}\frac{f(x)-g(x)}{g(x)}=0$. I am stuck right here!

Is there any need for further assumptions to carry out the proof from this side?

Answer 1

If $f(x)=g(x)+1$ and $\lim_{n\to a}g(x)=\infty$ then also $\lim_{n\to a}f(x)=\infty$, and: $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\left(1+\frac1{g(x)}\right)=1$$

But $f(x)-g(x)=1$ so there will be no convergence to $0$.