1
$\begingroup$

Suppose that $\lim_{x \to a^{-} } f(x) \lt \lim_{x \to a^{+} } f(x)$. Prove that there is some $\delta \gt 0$ such that $f(x) \lt f(y)$ whenever $x \lt a \lt y$ and $ |x-a| \lt \delta$ and $ |y-a| \lt \delta$.


My solution:

We know that $\exists ~\delta_1$ such that for every $\epsilon \gt 0$ $$ 0 \lt a -x \lt \delta_1 \implies |f(x) -L_1 | \lt \epsilon $$ Similarly, $$ 0 \lt x-a \lt \delta_2 \implies |f(x) - L_2| \lt \epsilon $$

Let $ \delta = min ( \delta_1, \delta_2)$, then we have $$ 0 \lt a -x \lt \delta \implies |f(x) -L_1 | \lt \epsilon \\ 0 \lt x-a \lt \delta \implies |f(x) - L_2| \lt \epsilon $$ Let's call the inputs greater than but within the $\delta$ of $a$ as $y$ $$ 0 \lt |x -a| \lt \delta \implies |f(x) -L_1 | \lt \epsilon ~~~~~~~~~~~~~~~x \lt a \\ 0 \lt |y-a| \lt \delta \implies |f(x) - L_2| \lt \epsilon~~~~~~~~~~~~~~~~~~y \gt a $$ By adding the two inequalities (involving $\epsilon$) we have $$ |f(x) - f(y) + L_2 -L_1| \lt 2 \epsilon \\ \text{ let $\epsilon = \frac{L_2 - L_1}{2}$, as we know $L_2 \gt L_1$} $$

$$|f(x) - f(y) + L_2 -L_1| \lt L_2 - L_1 \\ L_1 - L_2 \lt f(x) - f(y) + L_2 -L_1 \lt L_2 -L_1$$

$$2 (L_1 - L_2 ) \lt f(x) -f(y) \lt 0 \\ f(x) - f(y) \lt 0 \\ f(x) \lt f(y) $$

Is my solution correct and rigorous?

$\endgroup$

1 Answer 1

1
$\begingroup$

The final addition step doesn't seem correct. More simply, we need to assume a smaller value for $\varepsilon$ that is for example

$$\varepsilon \le \frac{L_2 - L_1}{3}$$

and then since $L_2>L_1$ we can conclude

$$|f(x) -L_1 | \lt \varepsilon \implies f(x)<\varepsilon+L_1=\frac{L_2 +2 L_1}{3}$$

$$|f(y) -L_2 | \lt \varepsilon \iff f(y)>-\varepsilon+L_2=\frac{2L_2 + L_1}{3}>f(x)$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .