Prove that if lim 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).

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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

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Solution: 

Since

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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,

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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

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with

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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.