- #1

issacnewton

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- 36

## Homework Statement

If ## \lim_{x \rightarrow a} f(x) = \infty## and ##\lim_{x \rightarrow a} g(x) = c ##, and if ##c>0## then prove that

##\lim_{x \rightarrow a} \left[ f(x)g(x)\right] = \infty~~ \text{if c > 0}##

## Homework Equations

Epsilon Delta definition of the limit

## The Attempt at a Solution

Let ##M>0## be arbitrary. Since ##c>0##, we have ##\frac{2M} c >0## and ##\frac c 2 > 0##. Since ##\lim_{x \rightarrow a} f(x) = \infty## , ##\exists ~ \delta_1 >0## such that $$\forall ~ x \in D(f)\left[ 0 <|x-a| < \delta_1 \rightarrow f(x) > \frac{2M} c \right]\cdots\cdots(1)$$ Similarly, 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 the functions ##f## and ##g##. Now let ##\delta = \text{min}(\delta_1, \delta_2) ## and let ##x_1 \in D(fg)## be arbitrary element. Then ##x_1 \in D(f)## and ##x_1 \in D(g)##. Also suppose ##0 <|x_1-a| < \delta ##. It follows from the definition of ##\delta## that ##0 <|x_1-a| < \delta_1 ## and ##0 <|x_1-a| < \delta_2 ##. So using the equations 1 and 2, it follows that ##f(x_1) > \frac{2M} c## and ##|g(x_1) - c| < \frac c 2##. So we have ##\frac c 2 < g(x_1) < \frac{3c} 2##. Since ##g(x_1) > \frac c 2 > 0##, we can deduce that ## f(x_1) g(x_1) > g(x_1) \frac{2M} c## But ##g(x_1) \frac{2M} c > \frac c 2 \bullet \frac{2M} c ##. So it follows that ##f(x_1) g(x_1) > \frac c 2 \bullet \frac{2M} c ##. Hence ## f(x_1) g(x_1) > M##. Since ##M>0## is arbitrary and ##x_1 \in D(fg)## is arbitrary, it follows that $$ \lim_{x \rightarrow a} \left[ f(x)g(x)\right] = \infty~~ \text{if c > 0}$$ Does this look correct ?

Thanks