Given e>0 find an interval I=(5,5+delta ),delta gt 0 such that if x lies in I, then sqrt [4](x-5)lt e What limit is being verified and what is its value? If x lies in the interval I=(5,5+8) then sqrt [4](x-5)lt e when 0lt delta lt square (Type an expression using eas the variable )

## Step 1<br />The problem is asking us to find an interval \(I=(5,5+\delta)\), where \(\delta > 0\), such that if \(x\) lies in \(I\), then \(\sqrt[4]{x-5} < e\). This is a limit problem, and we are verifying the limit of the function \(f(x) = \sqrt[4]{x-5}\) as \(x\) approaches 5.<br /><br />## Step 2<br />The limit we are verifying is \(\lim_{{x \to 5}} \sqrt[4]{x-5}\). We need to find a value for \(\delta\) such that for all \(x\) in the interval \(I=(5,5+\delta)\), the inequality \(\sqrt[4]{x-5} < e\) holds.<br /><br />## Step 3<br />To find the value of \(\delta\), we need to solve the inequality \(\sqrt[4]{x-5} < e\). This can be done by raising both sides of the inequality to the power of 4, which gives us \(x-5 < e^).<br /><br />## Step 4<br />Solving for \(x\), we get \(x < 5 + e^4\). This means that the interval \(I\) must be less than \(5 + e^4\).<br /><br />## Step 5<br />Therefore, the value of \(\delta\) must be less than \(5 + e^4 - 5\), which simplifies to \(\delta < e^4\).