Find a $\delta_1$ and a $\delta_2$ I have the majority done on this problem but the final step I'm having trouble understanding how to prove. I have put my proof so far in the answers. Can someone look at it and tell me if I'm showing it correctly and help me understand how to prove that $|g(f(x))-18|< \epsilon$?
The function $f(x)=\frac{11x-9}{4x}$ is continuous at $x=3$, $f(3)=2$, and the function $g(y)=y^2+7y$ is continuous at $y=2$.  Continuity of $g \circ f$ at $3$ follows from Theorem 3.4, and so for any $\epsilon >0$ there is a $\delta >0$ so that
$| g(f(x)) - g(f(3)) | < \epsilon$ for  all $x$ satisfying $| x-3 | < \delta$.
Given $\epsilon > 0$, determine a $\delta > 0$ by first calculating a suitable $\delta_1$ using the continuity of $g$ and then determining $\delta_2 > 0$ using the continuity of $f$ and this $\delta_1$.
 A: $g(y) = y^2 + 7y$ is continuous at $y = 2$. Continuity follows from theorem 3.4 for $g \circ f$ at 3, for any $\epsilon > 0$ there exists $\delta > 0$ so that $|g(f(x)) - g(f(3))|<\epsilon$ for all $x$ such that $|x - 3| < \delta$. We can show that
$|g(y)-g(2)| = |y^2 + 7y -18| = |y-2||y+9|$
Since $|y-2| < \epsilon = 1$ making $y < 3$ and for $|y+9| \le |y| + 9 \le 3+9 = 12$ implies that  $\delta_1 = \min(\frac{\epsilon}{12},1)$
We can now use $\delta_1$ as $\epsilon$ for $f$. We want show $\delta_2 = \frac{16\delta_1}{3}$ such that $|x-3|<\delta_2$ implies $|f(x) - 2|<\delta_1$. Then
$|f(x) - 2|=|\frac{11x-9}{4x}-2|=|\frac{3x-9}{4x}|=\frac{3|x-3|}{4|x|}$
Since $|x-3|<\epsilon = 1$ making $x<4$ and for $\frac{3}{4|x|} \le \frac{3}{16}$ making
$\delta_2 =\frac{16 \delta_1}{3}=\frac{16\min(\frac{\epsilon}{12},1)}{3}$ 
Now if $|x-2| < \frac{16\delta_1}{3}$ then $|g(f(x))-18|<\epsilon$
A: For any $\epsilon >0$, there exists $\delta_1 >0$ such that when $|f(x)-2|<\delta_1$, $|g(f(x)-g(2)|<\epsilon$;
And for this $\delta_1$, there exists $\delta_2 >0$ such that when $|x-3|<\delta_2$, $|f(x)-2|<\delta_1$.
So we have:

For any $\epsilon >0$,  there exists $\delta_2 >0$, such that when $|x-3|<\delta_2$, $|g(f(x)-g(2))|<\epsilon$.

