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Let $x_0$ be a real number and suppose that $f(x)$ and $g(x)$ are real valued functions defined for all $x>x_0$. If $\lim \limits_{x \to x_0+}f(x)=\infty$ and $\lim \limits_{x \to x_0+}g(x)=-\frac{1}{2}$. What do you expect the value of $\lim \limits_{x \to x_0+} f(x)g(x) $ to be? Confirm your guess by using the appropriate definition of a limit.

So I expect that the value of $\lim \limits_{x \to x_0+}f(x)g(x)=-\infty$

The definition of a limit is:

if $\forall \epsilon >0$ $ \exists \delta=\delta(\epsilon,x_0)$ such that $|f(x)-L|<\epsilon$ for all x $0< |x-x_0|<\delta$

I know that $\lim \limits_{x \to x_0+}f(x)g(x) = \lim \limits_{x \to x_0+}f(x)*\lim \limits_{x \to x_0+}g(x)$

I understand how to find the limits for $f(x)$ and $g(x)$, but how can I prove their product?

Is it as simple as saying that any constant multiplied by a divergent series is a divergent series?

What could be my value for $\delta$ that proves this to be true?

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  • $\begingroup$ Remember that $\infty$ is not a number, so you can not attempt to set $L=\infty$ in the definition of a limit. Rather, you must show that there is no limit for the function by proving the negative of the definition. $\endgroup$ – Richard P Oct 22 '15 at 2:41
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If $g(x) \to -1/2$ as $x \to x_{0}+$, then there is some $\delta_{1} > 0$ such that $0 < x-x_{o} < \delta_{1}$ only if $g(x) < \frac{1}{4} - \frac{1}{2} = \frac{-1}{4}$ and only if $f(x)g(x) < \frac{-1}{4}f(x)$; if $f(x) \to \infty$ as $x \to x_{0+}$, then, given any $M < 0$, there is some $\delta_{2} > 0$ such that $0 < x-x_{0} < \delta_{2}$ only if $f(x) > 4|M|$ and only if $$ \frac{-1}{4}f(x) < -|M| = M; $$ hence we have prove this: for every $M < 0$, we have $0 < x-x_{0} < \min \{\delta_{1}, \delta_{2} \}$ only if $f(x)g(x) < M$, which means that $f(x)g(x)$ diverges to the minus infinity as $x \to x_{0}+$.

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  • $\begingroup$ where did you get $\frac{1}{4}$ in "$g(x)<\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$" $\endgroup$ – Ryan T. Donnelly Oct 22 '15 at 3:14
  • $\begingroup$ never mind...it's because it's between 0 and $\frac{1}{2}$, right? $\endgroup$ – Ryan T. Donnelly Oct 22 '15 at 3:15
  • $\begingroup$ Yes; more precisely, it is chosen to make $g(x)$ negative. $\endgroup$ – Megadeth Oct 22 '15 at 3:23

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