I am not sure whether the method shown below is completely general or highly efficient, but it is a method. My recollection from school (back when we took the mastodon to school rather than a bus) is that these problems were posed without any fairly general method of solution being provided. So here is my method applied to your first problem. (I do not guarantee that my method is completely general or highly efficient. Moreover, I am sure it is not original with me.)
Step 1: Rewrite the function two ways: in terms of approaching the limiting argument (which is 1 in this example) from above and from below.
\(\displaystyle As\ x \rightarrow 1, f(x) = \dfrac{1}{x + 7} = \dfrac{1}{(1 + \kappa) + 7} = \dfrac{1}{8 + \kappa},\ provided\ 0 < \kappa.\) Approaching 1 from above.
\(\displaystyle As\ x \rightarrow 1, f(x) = \dfrac{1}{x + 7} = \dfrac{1}{(1 - \kappa) + 7} = \dfrac{1}{8 - \kappa},\ provided\ 0 < \kappa\ and\ \kappa \ne 8.\) Approaching 1 from below.
Now we can see that if x is almost 1, f(x) is positive, but if \(\displaystyle \kappa\) > 8, the second version of the function will be negative.
\(\displaystyle So\ 0 < \kappa < 8.\)
Step 2: Determine the absolute value of the difference between each revised version of the function and the proposed limit.
\(\displaystyle 0 < \kappa < 8 \implies \left|\dfrac{1}{8 + \kappa} - \dfrac{1}{8}\right| = \left|\dfrac{(1)(8) - (1)(8 + \kappa)}{8(8 + \kappa)}\right| = \left|\dfrac{- \kappa}{64 + 8\kappa }\right| = \dfrac{\kappa}{64 + 8 \kappa} > 0.\)
\(\displaystyle 0 < \kappa < 8 \implies \left|\dfrac{1}{8 - \kappa} - \dfrac{1}{8}\right| = \left|\dfrac{(1)(8) - (1)(8 - \kappa)}{8(8 - \kappa)}\right| = \left|\dfrac{\kappa}{64 - 8\kappa}\right| = \dfrac{\kappa}{64 - 8 \kappa} > 0.\)
Step 3: Determine which absolute value is the smaller (either by taking a difference or a ratio).
\(\displaystyle 0 < \kappa < 8 \implies \dfrac{\kappa}{64 - 8\kappa} \div \dfrac{\kappa}{64 + 8\kappa} = \dfrac{\kappa}{64 - 8\kappa} * \dfrac{64 + 8 \kappa}{\kappa} = \dfrac{64 + 8 \kappa}{64 - 8\kappa} > 1 \implies \dfrac{\kappa}{64 - \kappa} > \dfrac{\kappa}{64 + 8 \kappa}.\)
Step 4: Equate the smaller absolute value to epsilon (remembering epsilon is positive) and solve for kappa.
\(\displaystyle \dfrac{\kappa}{64 + 8 \kappa} = \epsilon \implies \kappa = 64\epsilon + 8\kappa\epsilon \implies \kappa - 8\kappa\epsilon = 64\epsilon \implies \kappa(1 - 8\epsilon) = 64\epsilon \implies \kappa = \dfrac{64\epsilon}{1 - 8\epsilon}\ if\ 0 < \epsilon < \dfrac{1}{8}.\)
So one possible answer is to set delta < kappa. But delta can be ANY positive number less than kappa, and the expression for kappa will not be easy to work with when we try to check our answer. This is a refinement but we can get an answer that is easier to use.
\(\displaystyle 0 < \epsilon < \dfrac{1}{8} \implies \kappa = \dfrac{64\epsilon}{1 - 8\epsilon} > \dfrac{64 \epsilon}{1 + 8\epsilon} > \dfrac{64\epsilon}{1 + 1} = 32\epsilon > 0.\)
An alternative answer is: \(\displaystyle 0 < \delta < 32\epsilon.\)
Step 5: Eliminate any restrictions on epsilon by finding delta that results in a value smaller than smallest restriction.
\(\displaystyle 0 \le \left|\dfrac{1}{8 + \delta} - \dfrac{1}{8}\right| < \dfrac{1}{8} \implies 0 \le \left|\dfrac{8 - (8 + \delta)}{8(8 + \delta)}\right| < \dfrac{1}{8} \implies 0 < \dfrac{\delta}{64 + 8\delta} < \dfrac{1}{8} \implies 0 < 8\delta < 64 + 8\delta \implies 0 < \delta.\)
\(\displaystyle 0 \le \left|\dfrac{1}{8 - \delta} - \dfrac{1}{8}\right| < \dfrac{1}{8} \implies 0 \le \left|\dfrac{8 - (8 - \delta)}{8(8 - \delta)}\right| < \dfrac{1}{8} \implies 0 < \dfrac{\delta}{64 - 8\delta} < \dfrac{1}{8} \implies 0 < 8\delta < 64 - 8\delta \implies 0 < \delta < \dfrac{64}{16} = 4.\)
\(\displaystyle THUS\ 0 < \delta < min\left(4,\ 32\epsilon\right).\)
Step 6: Check your work.
\(\displaystyle \dfrac{1}{8} \le \epsilon\ and\ 0 < \delta < 4 \implies (1 + 0) + 7 < (1 + \delta) + 7 < (1 + 4) + 7 \implies 8 < (1 + \delta) + 7 < 12 \implies \dfrac{1}{12} < \dfrac{1}{(1 + \delta) + 7} < \dfrac{1}{8} \implies\)
\(\displaystyle \dfrac{1}{12} - \dfrac{1}{8} < \dfrac{1}{(1+ \delta) + 7} - \dfrac{1}{8} < 0 \implies \left|\dfrac{1}{(1 + \delta) + 7} - \dfrac{1}{8}\right| < \dfrac{1}{8} - \dfrac{1}{12} = \dfrac{3 - 2}{24} = \dfrac{1}{24} < \dfrac{1}{8} \le \epsilon.\) That checks.
\(\displaystyle \dfrac{1}{8} \le \epsilon\ and\ 0 < \delta < 4 \implies - 4 < - \delta < 0 \implies (1 - 4) + 7 < (1 - \delta) + 7 < (1 - 0) + 7 \implies 4 < (1 - \delta) + 7 < 8 \implies\)
\(\displaystyle \dfrac{1}{8} < \dfrac{1}{(1 - \delta) + 7} < \dfrac{1}{4} \implies \dfrac{1}{8} - \dfrac{1}{8} < \dfrac{1}{(1 - \delta) + 7} - \dfrac{1}{8} < \dfrac{1}{8} \implies \left|\dfrac{1}{(1 - \delta) + 7} - \dfrac{1}{8}\right| < \dfrac{1}{4} - \dfrac{1}{8} = \dfrac{1}{8} \le \epsilon.\) That checks.
\(\displaystyle 0 < \epsilon < \dfrac{1}{8}\ and\ 0 < \delta < 32\epsilon \implies 8 + 0 < 8 + \delta < 8 + 32\epsilon \implies \dfrac{1}{8 + 32\epsilon} < \dfrac{1}{8 + \delta} < \dfrac{1}{8} \implies \dfrac{1}{8 + 32\epsilon} -\dfrac{1}{8} < \dfrac{1}{8 + \delta} - \dfrac{1}{8}< 0 \implies\)
\(\displaystyle \left|\dfrac{1}{(1 + \delta) + 7} -\dfrac{1}{8}\right| < \dfrac{1}{8} - \dfrac{1}{8 + 32\epsilon} = \dfrac{(1)(8 + 32\epsilon) - (1)8}{8(8 + 32\epsilon)} = \dfrac{4\epsilon}{8 + 32\epsilon} < \dfrac{4\epsilon}{8} < \epsilon.\) That checks.
\(\displaystyle 0 < \epsilon < \dfrac{1}{8}\ and\ 0 < \delta < 32\epsilon < 4 \implies - 4 < -32\epsilon < -\delta < 0 \implies 8 - 4 < 8 - 32\epsilon < 8 - \delta < 8 - 0\implies 4 < 8 - 32\epsilon < 8 - \delta < 8 \implies\)
\(\displaystyle \dfrac{1}{8} < \dfrac{1}{8 - \delta} < \dfrac{1}{8 -32\epsilon} \implies \dfrac{1}{8} - \dfrac{1}{8} < \left|\dfrac{1}{(1 - \delta) + 7} - \dfrac{1}{8}\right|< \dfrac{1}{1 - 32\epsilon} - \dfrac{1}{8} \implies\)
\(\displaystyle 0 < \left|\dfrac{1}{(1 - \delta) + 7} -\dfrac{1}{8}\right| < \dfrac{(1)(8) - (1)(8 - 32\epsilon)}{8(8 - 32\epsilon)} = \dfrac{4\epsilon}{8 - 32\epsilon} < \dfrac{4\epsilon}{4} = \epsilon.\) That checks.
