$|f(x)| \leq g(x) \forall x$, and $\lim _{x \rightarrow a} g(x)=0$, what is $\lim _{x \rightarrow a} f(x)$? What if $\lim _{x \rightarrow a} g(x)=5$?

Question

$$ \begin{aligned} &\text { Suppose }|f(x)| \leq g(x) \text { for all } x . \text { What can you conclude about } \lim _{x \rightarrow a} f(x) \text { if } \lim _{x \rightarrow a} g(x)=0 ?\\ &\text { What if } \lim _{x \rightarrow a} g(x)=5 ? \end{aligned} $$

Hint: remember that $|f(x)|\le g(x)$ is equivalent to $-g(x) \le f(x) \le g(x)$. — Greg Martin, Nov 01 '20 at 05:38
yes I know $ -g(x) \leq f(x) \leq g(x) $. But I don't know why that relates to the limits. — Jim Danger, Nov 01 '20 at 05:46
Re Greg Martin's comment, see https://en.wikipedia.org/wiki/Squeeze_theorem. — user2661923, Nov 01 '20 at 05:56

score 3 · Accepted Answer · answered Nov 01 '20 at 06:28

3

$|f(x)| \leq g(x)$ means that $$-g(x) \leq f(x) \leq g(x)$$

If $g(x)\to 0$, then $-g(x)\to 0$ too! Can we use the Sandwich theorem?

However, if $g(x)\to 5$, then $-g(x) \to -5$ and not much can be said about $\lim_{x\to a}f(x)$ in this case.

answered Nov 01 '20 at 06:28

stoic-santiago

13,705

Thanks. This makes sense. – Jim Danger Nov 01 '20 at 06:35

score 0 · Answer 2 · answered Nov 01 '20 at 06:05

0

HINT:

Can the sketch be useful in finding limit of $f(x)$?

answered Nov 01 '20 at 06:05

Narasimham

40,495

$|f(x)| \leq g(x) \forall x$, and $\lim _{x \rightarrow a} g(x)=0$, what is $\lim _{x \rightarrow a} f(x)$? What if $\lim _{x \rightarrow a} g(x)=5$?

2 Answers2