Question

Find the set of δ values that satisfy the formal definition of lim(x→4)⁡〖11/(g(x))〗=2.2 when given the value ε = 0.5, showing all work. Knowing that g(x)= x^2-x-7

Answer 1

\[\lim_{x\to4}\frac{11}{x^2-x-7}=2.2~~?\]

Answer 2

Nevermind, I solved it. It is using a delta epsilon proof (calc III) so it is intensive. Thanks Anyway

Answer 3

To find the proper \(\delta\), assume \(\left|\dfrac{11}{x^2-x-7}-\dfrac{11}{5}\right|<\epsilon\). Then,
\[\begin{align*}\left|\dfrac{11}{x^2-x-7}-\dfrac{11}{5}\right|&=11\left|\dfrac{5-(x^2-x-7)}{x^2-x-7}\right|\\
&=11\left|\frac{x^2-x-12}{x^2-x-7}\right|\\
&=11\left|\frac{(x-4)(x+3)}{x^2-x-7}\right|\\
&=11\left|x-4\right|\left|x+3\right|\left|\frac{1}{x^2-x-7}\right|
\end{align*}\]
Agree to set \(\delta\le1\), then if \(0<|x-4|<\delta\), you have that
\[\begin{align*}
|x-4|&<1\\
-1<x-4&<1\\
6<x+3&<8\\
|x+3|&<8
\end{align*}\]
So,
\[11\left|x-4\right|\left|x+3\right|\left|\frac{1}{x^2-x-7}\right|<88\left|x-4\right|\left|\frac{1}{x^2-x-7}\right|\]
For the quadratic factor, I think completing the square would be your best bet:
\[x^2-x-7=x^2-x+\frac{1}{4}+\frac{27}{4}=\left(x-\frac{1}{2}\right)^2+\frac{27}{4}\]
So if \(\delta\le1\), you get
\[\begin{align*}
|x-4|&<1\\
-1<x-4&<1\\
\frac{5}{2}<x-\frac{1}{2}&<\frac{9}{2}\\
\left|x-\frac{1}{2}\right|&<\frac{9}{2}\\
\left(x-\frac{1}{2}\right)^2&<\frac{81}{4}\\
\left(x-\frac{1}{2}\right)^2+\frac{27}{4}&<\frac{108}{4}=27\\
\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{27}{4}}&>\frac{1}{27}
\end{align*}\]
So,
\[88\left|x-4\right|\left(\frac{1}{27}\right)<88\left|x-4\right|\left|\frac{1}{x^2-x-7}\right|<\epsilon\]
or simply
\[\frac{88}{27}\left|x-4\right|<\epsilon\\
|x-4|<\frac{27\epsilon}{88}\]
Which would mean you would have to set \(\delta=\min\left\{1,\dfrac{27\epsilon}{88}\right\}\). So if \(\epsilon=\dfrac{1}{2}\) is given, you would pick \(\delta=\min\left\{1,\dfrac{44}{27}\right\}=1\). Now, I get the feeling I might be missing something in the proof, but I'm comfortable with it overall.

Answer 4

... Just noticed that you have your answer... Oh well :)