Prove that if \[\lim_{h\to 0}\,g(h)=0\] and $g(h)\neq 0$ for some interval $(-\bar{\delta},0)\cup (0,\bar{\delta})$ where $\bar{\delta}\gt 0$ then \[ \lim_{h\to 0}\,g(h)=\lim_{g(h)\to 0}\,g(h). \]

Primary tabs

Lesson Parent: 
Function Limits III: The Gory Details
Problem: 

Prove that if \[\lim_{h\to 0}\,g(h)=0\] and $g(h)\neq 0$ for some interval $(-\bar{\delta},0)\cup (0,\bar{\delta})$ where $\bar{\delta}\gt 0$ then
\[
\lim_{h\to 0}\,g(h)=\lim_{g(h)\to 0}\,g(h).
\]

Answer: 

\[ \lim_{h\to 0}\,g(h)=\lim_{g(h)\to 0}\,g(h). \]

SAMPLE SOLUTION

All of our problems have fully worked out, logical solutions. We connect every solution to its underlying lesson so you can understand the concepts needed to solve any problem and reinforce your learning by practicing with similar problems. All of the formulas are available to our subscribers, but here is a sample of the problem solution you will see when you log in:



Solution: 

Since

<Sign in to see all the formulas>


we know that for every every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that if <Sign in to see all the formulas> then <Sign in to see all the formulas>.

Furthermore, if we are at an $h$ such that <Sign in to see all the formulas> then <Sign in to see all the formulas>.

Now putting it together we know that for every <Sign in to see all the formulas> there exists a <Sign in to see all the formulas> such that if <Sign in to see all the formulas> then <Sign in to see all the formulas>. Therefore,

<Sign in to see all the formulas>

The point of this problem is that even though we know that $g(h)$ goes to zero as $h$ goes to zero, we can only make the substitution of

<Sign in to see all the formulas>

with

<Sign in to see all the formulas>

when $g(h)$ is not zero. This is because our limit definition requires we not actually get to the point we are wanting the limit at. In other words, $g(h)$ must get close to zero without actually being zero.