So first we need to define what it means to be a limit. There ar etwo major definitions, which are essentially the same. I have used both, the first from Calc 1 to calc 3 and the latter from the proofs class i'm doing now. I will prove the first question using both methods, so you can see how they are similar and different, and the last by the latter definition only. Here are the definitions:

Definition 1:

Let be a function defined on an open interval containing (except possibly at ) and let be a real number. means that

for each real there exists a real such that for all where we have .

Definition 2:

We say the limit of a sequence (Sn) is a real number, s, provided that for each e > 0 there exists a natural number N such that n>N implies |Sn - S| < e

Note: e means epsilon. d means delta

lim (KSn) = KSProof (By definition 1):The result is obvious for k = 0. Assume k is not 0. Let limSn = S. So for each e>0 there exists a d>0 such that 0<|x - c|<d implies |Sn - S|<e/|k|.

Then 0<|x - c|<d implies |kSn - kS| <or= |k||Sn - S| < |k|.e/|k| = e

So we have for each e>0, there is a d>0 such that 0<|x - c|<d implies |kSn - kS|< e. Thus the limkSn = kS

QED

Alternate Proof (By definition 2):The result is obvious for k = 0, therefore, let us assume k is not 0.

Let e>0. Since limSn = S, there exists an N such that n>N implies |Sn - S|< e/|k|

So n>N implies |kSn - kS| <or= |k||Sn - S| < |k|.e/|k| = e

Since n>N implies |kSn - kS| < e, limkSn = kS.

QED

lim (K + Sn) = K + SProof (By Definition 2):The result is obvious for k = 0, so let's assume k is not 0. Let the limSn = S. This means for any e > 0, there exists an N such that n > N implies |Sn - S| < e

Then n>N implies |(K + Sn) - (K + S)| = |Sn - S + K - K| = |Sn - S| < e

Since n>N implies |(K + Sn) - (K + S)| < e, lim(K + Sn) = K + S

QED

The proof by definition 1 is very similar as you can see in the first question I did. You should be able to reproduce this result using that definition if that's the one you learned in class.