# Suppose that lim_(xrarrc) f(x) = 0 and there exists a constant K such that ∣g(x)∣ ≤ K " for all " x nec in some open interval containing c. Show that lim_(x→c) (f(x)g(x)) = 0?

##### 2 Answers
May 8, 2016

If $K = 0$ then $g \left(x\right) = 0$ for all $x \ne c$ in some open interval containing $c$, and so ${\lim}_{x \to c} \left(f \left(x\right) g \left(x\right)\right) = {\lim}_{x \to c} 0 = 0$. Then, for the remainder of the explanation, we will assume $K > 0$.

Using the $\epsilon - \delta$ definition of limits, we can say that ${\lim}_{x \to c} \left(f \left(x\right) g \left(x\right)\right) = 0$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $0 < | x - c | < \delta$ implies $| f \left(x\right) g \left(x\right) - 0 | < \epsilon$.

To show that this is the case, first we let $\epsilon > 0$ be arbitrary.

Now, as $\frac{K}{\epsilon} > 0$ and ${\lim}_{x \to c} f \left(x\right) = 0$, we know by the above definition there exists some $\delta ' > 0$ such that $| g \left(x\right) | < K$ for $x \in \left(c - \delta ' , c + \delta '\right) \setminus \setminus \setminus \left\{c\right\}$ and if $0 < | x - c | < \delta '$ then $| f \left(x\right) - 0 | < \frac{\epsilon}{K}$.

Let $\delta = \delta '$. Then, for $0 < | x - c | < \delta$, noting that as $| x - c | > 0$ we know that $x \ne c$, we have

$| f \left(x\right) g \left(x\right) - 0 | = | f \left(x\right) g \left(x\right) |$

$= | g \left(x\right) | \cdot | f \left(x\right) |$

$\le K | f \left(x\right) | \text{ }$ (as $x \in \left(c - \delta ' , c + \delta '\right) \setminus \setminus \setminus c \implies K \ge | g \left(x\right) |$)

$< K \left(\frac{\epsilon}{K}\right) \text{ }$ (as $0 < | x - c | < \delta = \delta ' \implies | f \left(x\right) | < \frac{\epsilon}{K}$)

$= \epsilon$

Thus, as we have demonstrated the existence of such a $\delta$ for an arbitrary $\epsilon$, we may conclude that ${\lim}_{x \to c} \left(f \left(x\right) g \left(x\right)\right) = 0$.

May 8, 2016

See below for a proof using the squeeze theorem.

#### Explanation:

${\lim}_{x \rightarrow c} f \left(x\right) = 0$ implies ${\lim}_{x \rightarrow c} \left\mid f \left(x\right) \right\mid = 0$

$\left\mid g \left(x\right) \right\mid \le K$ if and only if $- K \le g \left(x\right) \le K$ for $x$ sufficiently close to $c$ except perhaps at $x = c$.

$- K \le g \left(x\right) \le K$ and $\left\mid f \left(x\right) \right\mid \ge 0$ implies

$- \left\mid f \left(x\right) \right\mid K \le \left\mid f \left(x\right) \right\mid g \left(x\right) \le \left\mid f \left(x\right) \right\mid K$.

Now, since ${\lim}_{x \rightarrow c} \left(- \left\mid f \left(x\right) \right\mid K\right) = 0 = {\lim}_{x \rightarrow c} \left(\left\mid f \left(x\right) \right\mid K\right)$,

we can conclude that ${\lim}_{x \rightarrow c} \left(\left\mid f \left(x\right) \right\mid g \left(x\right)\right) = 0$.