In Standard Analysis, Does $$\lim_{x \to c}\frac{f(x)}{g(x)}=\frac{\lim_{x \to c}f(x)}{\lim_{x \to c}g(x)}$$ Given that $$\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0$$ Yes, we can get the computed answer for Limit through L'Hospital's rule but what resolution do we have upon this situation about equivalence. For below function, we can get L = 4 by L'Hospital's rule, but can we say that $$(\lim_{x \to c}\frac{f(x)}{g(x)}=4)=(\frac{\lim_{x \to c}f(x)}{\lim_{x \to c}g(x)}=\frac{0}{0})?$$What do we say of the 2 sides together i.e relation b/w them ? Any Equivalence b/w the 2 sides ? In contrast, equality is indeed possible when the 2 individual limits are non zero.
In Non Standard Analysis (Robinson's Hyperreal Framework), does $$st(\frac{a}{b})=\frac{st(a)}{st(b)}?$$ if $st(a)=0,st(b) = 0$
E.g $$f(x)=\frac{x^2-4}{x-2}\\ \lim_{x \to 2} \ f(x) = 4$$ Non standard way, let $$g(x) = x^2-4\\ h(x) = x-2\\$$ dx is an infinitesimal $$L=st(\frac{g(2+dx)}{h(2+dx)})=st(4+dx)=4$$ Noting that $$\frac{st(g(2+dx))}{st(h(2+dx))}=\frac{M}{N}=\frac{0}{0}=\frac{\lim_{x \to 2}g(x)}{\lim_{x \to 2}h(x)}$$
What can we say about the L, i.e = $$(\lim_{x \to c}\frac{f(x)}{g(x)}=4)$$, and the 0/0 situation. Any equivalence b/w them ?
In cases where, M and N are non zero,$$st(\frac{a}{b})=\frac{st(a)}{st(b)}=\frac{M}{N}$$ is permissible.
The only case where $$st(\frac{a}{b})\neq\frac{st(a)}{st(b)}$$ is where $$st(b)=0$$
Please Ref - Table 3.3.1, Pg-121, Elementary Calculus: An Infinitesimal Approach, Howard Jerome Keisler
$0/0$ does not exist. We define $+$ and $\times$. Then we define $a-b$ as the unique $c$ such that $a=b+c$, equivalently that $a-b=a+b'$ where $b'$ is the unique additive inverse of $b$, that is $b+b'=0$. Similarly we define $a/b$ as the unique $d$ such that $a=bd$ IF such a unique $d$ exists, equivalently that $a/b=ab''$ if $b''$ is the unique multiplicative inverse of $b$, that is $bb''=1$, IF such a unique $b''$ exists. BUT if $b=0$ then no such $b''$ (unique or not) exists.