Delta Epsilon Limits Proving: a+L and a*L

[Shelby]

Homework Statement

Given that lim f(x)=L as x approaches a , prove that lim (x+f(x))=a+L as x approaches ahttps://www.physicsforums.com/attachments/9630. Your proof cannot assume that the limit of a sum of two functions is the sum of their individual limits. You must use the delta-epsilon definition of limit in your proof.
and
Given that lim f(x)=L as x approaches a , prove that lim x*f(x)=aL as x approaches a

Homework Equations

The Attempt at a Solution

attempt is in the attatchment

 

[VietDao29]

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:
$$0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$$What 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? :)

 

[Shelby]

Thank you, I do have one question

Why did you choose to divide epsilon 1 by 2?

 

[mjsd]

because eventually you want to sum up two of them.. and 1/2 +1/2 =1 gives you just one epsilon... and this matches the target you want to reach. that's all

 

[VietDao29]

Why did you choose to divide epsilon 1 by 2?

Since you want to prove that:
$$|[x + f(x)] - [a + L]| = |(x - a) + [f(x) - L]| < |x - a| + |f(x) - L| < \epsilon_1$$, i.e, the sum of |x - a|, and |f(x) - L|, so it's natural to choose the sum of $$\frac{\epsilon_1}{2}$$, and $$\frac{\epsilon_1}{2}$$, as $$\frac{\epsilon_1}{2} + \frac{\epsilon_1}{2} = \epsilon_1$$.
Then assign $$\delta, \ \delta_1 , \ \epsilon$$ wisely, so that: $$|x - a| < \frac{\epsilon_1}{2}$$, and $$|f(x) - L| < \frac{\epsilon_1}{2}$$

 

[pjallen58]

I think I understand the process shown for the above proof but if someone could provide the second part it would improve my understanding. Given that lim f(x)=L as x approaches a , prove that lim x*f(x)=aL as x approaches a using the delta epsilon definition of a limit. This would be much appreciated. Thanks.

 

[jacobrhcp]

just try it yourselves! Use the method and I'm confident it will work out!