- #1
fmam3
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Homework Statement
If [tex]f[/tex] is defined on [tex](a,b)[/tex], [tex]\lim_{x \to a^{+}} |f(x)| = +\infty[/tex] and [tex]f[/tex] is decreasing on [tex](a,b)[/tex], show that [tex]\lim_{x \to a^{+}} f(x) = +\infty[/tex].
Homework Equations
The Attempt at a Solution
If [tex]\lim_{x \to a^{+}} |f(x)| = +\infty[/tex], then [tex]\forall M > 0, \exists \delta > 0[/tex] such that [tex]\forall x \in (a,b)[/tex] and [tex]a < x < a + \delta[/tex], we have [tex]|f(x)| > M[/tex], which implies we either have [tex]f(x) > M[/tex] or [tex]f(x) < -M[/tex]. It suffices to show that [tex]f(x) < -M[/tex] does not hold.
For contradiction, suppose that [tex]f(x) < -M[/tex] for [tex]x \in (a,b) \cap (a, a + \delta)[/tex]. Consider a sequence of real numbers [tex](x_n)[/tex] such that [tex]x_n \in (a,b) \cap (a, a + \delta)[/tex] and [tex]x_{n+1} > x_n[/tex] and [tex]\lim_{n \to \infty}x_n = a[/tex] for [tex]\forall n \in \mathbb{N}[/tex]. Since [tex]f[/tex] is decreasing, [tex]x_{n+1} > x_n[/tex] implies [tex]f(x_n) \geq f(x_{n+1})[/tex]. Thus, we have that [tex]-M > f(x_n) \geq f(x_{n+1})[/tex]...
But this is the part where I got stuck... any help appreciated! Thanks in advance!