1
$\begingroup$

Suppose the limits $L_1 = \lim_{x \rightarrow a^+}f_1(x)$ and $L_2 = \lim_{x \rightarrow a^+}f_2(x)$ exist. Show that if $f_1(x) \leq f_2(x)$ for all x in some interval (a,b), then $L_1 \leq L_2$.

The proof I've seen just shows that since $f_1(x) \leq f_2(x)$ and since $f_1(x) \rightarrow L_1$ and $f_2(x) \rightarrow L_2$ this shows that $L_1 \leq L_2$. But I don't see how this proves it at all. The proof might be using something like squeeze theorem to prove it, but I want to understand how to prove it directly.

$\endgroup$
2
$\begingroup$

Let $g(x) = f_2(x) - f_1(x)$. Then $g(x) \geqslant 0$ for $x \in (a,b)$ and and $\lim_{x \to a+} g(x) = L = L_2 - L_1$.

Assume $L < 0$. For any $\epsilon > 0$ there exists $\delta > 0$ such that $g(x) < L + \epsilon$ if $a < x < a+ \delta$. Choose $\epsilon = -L/2$. This implies $g(x) < L/2 < 0$ if $a < x < a + \delta$.

Hence $f_2(x) < f_1(x)$ at some point in the interval -- a contradiction.

Therefore, $L = L_2 - L_1 \geqslant 0$ and $L_2 \geqslant L_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.