A) If () = () when ≠ then lim → () = lim → (), provided the limits exist
B) If lim →5 () = 0 and lim →5 () = 0 then lim →5 ()/() does not exist
Respectively my answers are true and then false, but I don't believe I am correct. I'm fairly new to limits and am struggling with the concept, a lil help would be nice, thank you!
On
B is false because when both $f$ and $g$ have limit 0, L'Hospital's Rule can be applied to $f/g$. For example, $\sin x$ and $x$ both go to 0 as $x \rightarrow 0$, but by L'Hospital's Rule, $\sin x/x$ goes to 1 as $x \rightarrow 0$. For A, suppose $\lim_{x \rightarrow a} f(x) = F$ and $\lim_{x_ \rightarrow a} g(x) = G$. By definition, given $\epsilon > 0$, there is $\delta_f > 0$ (likewise, $\delta_g > 0$) such that when $0 < |x - a| < \delta_f$ (likewise, $0 < |x - a| < \delta_g$), we have $|f(x) - F| < \epsilon/2$ (likewise, $|g(x) - G| < \epsilon/2$). Set $\delta = \min \lbrace \delta_{f}, \delta_{g} \rbrace$, then both inequalities are satisfied when $0 < |x - a| < \delta$. Since $f(x) = g(x)$ for all $x \not = a$, $|f(x) - F - [g(x) - G]| = |f(x) - g(x) - F + G| = |G - F|$ for all such $x$. Now for $0 < |x - a| < \delta$, $|f(x) - F - g(x) - G| \leq |f(x) - F| + |g(x) - G| < \epsilon/2 + \epsilon/2 = \epsilon$ (by the triangle inequality). Combining these two results yields $|G - F| < \epsilon$. Since the choice of $\epsilon$ is arbitrary, we must have $|G - F| = 0$, or $F = G$.
Your answers are correct. For B) you can take $f(x)=g(x)=x-5$ to show that the limit can exist.