Next, the negative must be distributed to the “a” and to the “L” of (a + L) and an L must be substituted for f(x) because it is given that the \lim_{x \rightarrow a} f(x) = L
|x + L - a - L|...
That part is wrong, you cannot sub L for f(x) like that, it's not correct.
Now, back at the start, the problem gives that:
\lim_{x \rightarrow a} f(x) = L, and you must use
this to prove: \lim_{x \rightarrow a} [x + f(x) ] = a + L
\lim_{x \rightarrow a} f(x) = L means that, for any arbitrary small \epsilon, there exists a \delta, such that:
0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilonWhat we should prove is:
For any arbitrary small \epsilon_1, there exists a \delta_1, such that:
0 < |x - a| < \delta_1 \Rightarrow |[x + f(x)] - [a + L]| < \epsilon_1
We will deal with this part:
|[x + f(x)] - [a + L]| = |(x - a) + [f(x) - L]| < |x - a| + |f(x) - L|Now we choose, \epsilon = \frac{\epsilon_1}{2}.
Now, there should exists a \delta, such that: 0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon = \frac{\epsilon_1}{2}
Choose \delta_1, so that: \delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right)
\delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right) \Rightarrow \delta_1 \leq \delta, so we have:
0 < |x - a| < \delta_1 \leq \delta \Rightarrow |f(x) - L| < \epsilon = \frac{\epsilon_1}{2}
\delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right) \Rightarrow \delta_1 \leq \frac{\epsilon}{2}Now, if we have:
0 < |x - a| < \delta_1 \Rightarrow |[x + f(x)] - [a + L]| < |x - a| + |f(x) - L| < \delta_1 + \epsilon < \frac{\epsilon_1}{2} + \frac{\epsilon_1}{2} = \epsilon_1, so:
\lim_{x \rightarrow a} [x + f(x) ] = a + L (Q.E.D)
Is there any where unclear?
Can you do the same to part II? :)
