Question

Prove that if lim x→c

​
f(x)=0 and ∣g(x)∣≤M for a fixed number M and all x

=c, then lim x→c
​
[f(x)g(x)]=0. x≤f(x)g(x)≤
lim x→c
​
x) )≤lim x→c
​
[f(x)g(x)]≤lim x→c
​
(x)
x≤lim x→c
​
[f(x)g(x)]≤x
0≤lim x→c
​
[f(x)g(x)]≤0
​
Therefore, lim x→c
​
[f(x)g(x)]=

Answer 1

We have proven that if
`\lim _(x \rightarrow c) f(x) g(x)=0` under the given conditions.

To prove the statement
`\lim _(x \rightarrow c)[f(x) g(x)]=0` under the given conditions,

we will use the Squeeze Theorem.

The Squeeze Theorem states that if
`h(x) \leq f(x) \leq g(x)` for all x in some open interval containing c, except possibly at c itself, and:

If
`\lim _(x \rightarrow c) h(x)=`
`\lim _(x \rightarrow c) g(x)=L`

then,
`\lim _(x \rightarrow c) f(x)=L`.

Given that
`\lim _(x \rightarrow c) f(x)=0 \text { and }|g(x)| \leq M` for some fixed number M, we can set up the following inequalities:

`-M \leq g(x) \leq M`

Multiply both sides of the above inequality equation by f(x):

`-M \cdot f(x) \leq f(x) g(x) \leq M \cdot f(x)`

Now, take the limit as x approaches c for each part:

`\begin{aligned}& \lim _(x \rightarrow c)(-M \cdot f(x))=-M \cdot \lim _(x \rightarrow c) f(x)=-M \cdot 0=0 \\& \lim _(x \rightarrow c)(M \cdot f(x))=M \cdot \lim _(x \rightarrow c) f(x)=M \cdot 0=0\end{aligned}`

By the Squeeze Theorem, since
`-M \cdot f(x) \leq f(x) g(x) \leq M \cdot f(x)` and both the lower and upper bounds approach 0 as x approaches c, it follows that:

`\lim _(x \rightarrow c) f(x) g(x)=0`

Complete Question: