Does $\lim_{x\to0} \frac{f(x)}{g(x)}=1$ imply that $\lim_{x\to0}f(x)-g(x)=0$? On an intuitive level, it would seem that
$$
\lim_{x\to0} \frac{f(x)}{g(x)}=1\implies\lim_{x\to0}f(x)-g(x)=0
$$
If the quotient of two quantities is approaching $1$, then the two quantities would have to be getting closer together. However, I'm not quite sure if the above statement is true for all functions $f,g$. More generally, is the following statement true?
$$
\lim_{x\to c} \frac{f(x)}{g(x)}=1\iff\lim_{x\to c}f(x)-g(x)=0
$$
 A: Let consider as counterexample
$$f(x)=\ln x+\frac1{x}, \quad g(x)=\frac1{x}$$
such that
$$\frac{f(x)}{g(x)} =x\ln x+1 \to 0+1=1$$
but
$$f(x)-g(x)=\ln x \to -\infty$$
A: Were it so easy!
The Prime Number Theorem says that if $\pi(x)$ counts the number of primes less than or equal to $x$,
$$
\lim_{x\to\infty} \frac{\pi(x)}{x/\log(x)}=1
$$If your claim were true, then eventually, say for some $M\in\mathbb{R}$, $|\pi(x)-x/\log(x)|<1/2$ for $x\ge M$. But since $\pi(x)$ is clearly an integer, this would give a closed-form for $\pi(x)$ (provided we knew $M$). As of this writing, however, such a formula is not known.
A: Take $f(x)=g(x) +h(x) $ such that $\lim h(x) /g(x) =0$ Take any fast growing function $g$ and take any slower function $h$ that satisfies the property.
A simple example is $$g(x) = \frac1{(x-c)^2},h(x)=1$$
For any $c$
A: $$
f(x) = \frac{1}{x} + \frac{1}{\sqrt{x}}\ ,\qquad g(x) = \frac{1}{x}
$$
$$
\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{\frac{1}{x} + \frac{1}{\sqrt{x}}}{\frac{1}{x}} = \lim_{x \to 0} \frac{1 + \sqrt{x}}{1} = 1
$$
$$
\lim_{x \to 0} (f(x) - g(x)) = \lim_{x \to 0} (\frac{1}{\sqrt{x}}) = \infty
$$
A: According to you algebraical transformations u have to do following steps:
$$ lim_{x\to 0}\frac{f(x)}{g(x)}=1 $$
$$\frac{\lim_{x\to 0}f(x)}{\lim_{x\to 0}g(x)}=1$$, then u multiplied both sides by $\lim_{x\to 0}g(x)$ and finally get:
$$\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x) $$
so it is the same as:
$$\lim_{x\to 0}f(x)-\lim_{x\to 0}g(x)=0 $$
$$\lim_{x\to 0}(f(x)-g(x))=0$$
The problem is when we split limit of a quotient into a quotient of a limit,and same is when we substract limits. We can do it under an assumption that:$$\lim_{x\to 0}f(x)=g_{1} $$ and $$\lim_{x\to 0}g(x)=g_{2}\ne 0 $$ where $-\infty\lt g_{1},g_{2}\lt\infty $ ,
Example, . Remember: while calculating limits, first think you do is to simplify expression inside the limit.
But, please correct me if I am wrong.
A: Many counterexamples have been provided, but there are simpler (an more illuminating, I think)
$$f(x) = \frac1{x} \quad g(x) = 1 + \frac1{x}$$
A: $f(x) = x^n+x^{n-1}, g(x) = x^n$
shows no
since
$\dfrac{f(x)}{g(x)}
=1+\dfrac1{x}
\to
1$.
