Question

I'm looking at Find the set of δ values that satisfy the formal definition of lim(11/g(x))=2.2 as x approaches 4 when given the value ε = 0.5 and g(x) = x2 – x – 7, showing all work. Can I say, if lim as X approaches 4 of 11/gx = 2.2, the lim of x approaches g9x) / 11 = 1/2.2...from there I ended up with abs (x-4) = 11ε/K. the process i showed was if abs(X-4)(X+3)/11<.5, then let K = X-3, to give abs (x-4)K <5.5 to give abs (x-4) <5.5/k. Since there is no K to satisfy all values of X, I continued with the formula and looked at and set abs (x-4) < 1 which gives an upper limit of 5. But since K = X+3, then K = 8.......thi is where i lost the proof....
Now I'm stuck! I know I'm close!

Answer 1

I would proceed with the limit's proof:
Given \(\epsilon>0\), find \(\delta\) such that \(0<|x-4|<\delta\) implies \(\left|\dfrac{11}{x^2-x-7}-2.2\right|<\epsilon\).
\[\begin{align*}\left|\frac{11}{x^2-x-7}-2.2\right|&=\left|\frac{11}{x^2-x-7}-\frac{11}{5}\right|\\
&=11\left|\frac{1}{x^2-x-7}-\frac{1}{5}\right|\\
&=11\left|\frac{5-\left(x^2-x-7\right)}{5\left(x^2-x-7\right)}\right|\\
&=\frac{11}{5}\left|\frac{x^2-x-12}{x^2-x-7}\right|\\
&=\frac{11}{5}\left|\frac{x+3}{x^2-x-7}\right||x-4|
\end{align*}\]
Agree to set \(\delta\le\dfrac{1}{2}\) (I started off with \(\delta\le1\), but I ran into problems. Putting a greater restriction on \(\delta\) seemed to help.) Then you have that
\[\begin{align*}|x-4|<\frac{1}{2}~~\Rightarrow~~-\frac{1}{2}&<x-4<\frac{1}{2}\\
\color{red}{\frac{13}{2}}&\color{red}{<x+3<\frac{15}{2}}\\\\
\frac{7}{2}&<x<\frac{9}{2}\\
3&<x-\frac{1}{2}<4\\
9&<\left(x-\frac{1}{2}\right)^2<16\\
\frac{7}{4}&<\left(x-\frac{1}{2}\right)^2-\frac{29}{4}<\frac{35}{4}\\
\frac{7}{4}&<x^2-x-7<\frac{35}{4}\\
\color{red}{\frac{4}{35}}&\color{red}{<\frac{1}{x^2-x-7}<\frac{4}{7}}
\end{align*}\]
Multiplying the red inequalities gives us
\[\frac{26}{35}<\frac{x+3}{x^2-x-7}<\frac{30}{7}~~\Rightarrow~~\left|\frac{x+3}{x^2-x-7}\right|<\frac{30}{7}\]
So, this gives us
\[\begin{align*}\frac{11}{5}\left|\frac{x+3}{x^2-x-7}\right||x-4|<\frac{11}{5}\cdot\frac{30}{7}|x-4|&<\epsilon\\
|x-4|&<\frac{7}{66}\epsilon
\end{align*}\]
All this work to show that you would choose \(\delta=\min\left\{\dfrac{1}{2},\dfrac{7}{66}\epsilon \right\}\).