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## Homework Statement

If ##\lim_{x \rightarrow a} f(x) =\infty## and ## \lim_{x \rightarrow a} g(x) = c##, then prove that $$\lim_{x \rightarrow a} \left[ f(x)g(x)\right] = -\infty ~\text{ if c < 0 }$$

## Homework Equations

Epsilon Delta definition

## The Attempt at a Solution

Let ##N<0## be arbitrary. Since ##c<0##, we have ##\frac{2N} c >0##. Since ##\lim_{x \rightarrow a} f(x) =\infty##, ##\exists~ \delta_1>0## such that $$\forall ~ x\in D(f)\left[ 0<|x-a| <\delta_1 \rightarrow f(x) > \frac{2N} c \right]\cdots\cdots(1)$$ Also ##-\frac c 2 >0##, and since ## \lim_{x \rightarrow a} g(x) = c##, ##\exists ~ \delta_2 >0 ## such that $$\forall ~x \in D(g) \left[ 0<|x-a| <\delta_2 \rightarrow |g(x) - c| < -\frac c 2 \right]\cdots\cdots(2)$$ Here ##D(f)## and ##D(g)## are the domains of functions ##f## and ##g##. Let ##\delta = \text{min}(\delta_1, \delta_2)## and let ##x_1 \in D(fg)## be arbitrary, where ##D(fg)## is the domain of function ##fg##. So we have ##x_1 \in D(f)## and ##x_1 \in D(g)##. Also suppose ##0<|x_1-a| <\delta ##. It follows that ##0<|x_1-a| <\delta_1## and ## 0<|x_1-a| <\delta_2##. So now using equations ##(1)## and ##(2)##, we deduce that ##f(x_1) > \frac{2N} c ## and ##|g(x_1)-c| < -\frac c2## It follows that ##g(x_1) < \frac c 2##. Since ##f(x_1) > \frac{2N} c > 0##, we get ## f(x_1)g(x_1) <f(x_1) \frac c 2##. But as ##f(x_1) > \frac{2N} c## , we have ##f(x_1)\frac c 2 < N##. It follows that ##f(x_1)g(x_1) < N##. As ##N<0## is arbitrary, and ##x_1 \in D(fg)## is arbitrary, from the definition it follows that $$\lim_{x \rightarrow a} \left[ f(x)g(x)\right] = -\infty ~\text{ if c < 0 }$$ Please check my proof. Thanx