Let {Xn}, {yn} be real sequences and b,x,y be real numbers.

Prove (using limit theorems)

If Xn goes to X and Yn goes to Y as n goes to infinity, then
i)lim (Xn + Yn) = X+Y
ii)lim (bXn) = bX
iii) lim (XnYn) = XY
(all limits n going to infinity)

Thanks

2. Originally Posted by taypez
Let {Xn}, {yn} be real sequences and b,x,y be real numbers.

Prove (using limit theorems)

If Xn goes to X and Yn goes to Y as n goes to infinity, then
i)lim (Xn + Yn) = X+Y
Let $\lim \ x_n = x$ and $\lim \ y_n = y$.
This means,
$|x_n - x|< \frac{\epsilon}{2} \mbox{ where }n>N_1$
$|y_n-y|<\frac{\epsilon}{2} \mbox{ where }n>N_2$
Thus,
$|x_n - x|< \frac{\epsilon}{2} \mbox{ and }|y_n-y|< \frac{\epsilon}{2} \mbox{ where }n>N \mbox{ where }N=\max\{N_1,N_2\}$.

Thus,
$|x_n+y_n - x - y| \leq |x_n - x|+|y_n-y| < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon \mbox{ for }n>N$.
Q.E.D.

3. Originally Posted by taypez
Let {Xn}, {yn} be real sequences and b,x,y be real numbers.

Prove (using limit theorems)

If Xn goes to X and Yn goes to Y as n goes to infinity, then
i)lim (Xn + Yn) = X+Y
ii)lim (bXn) = bX
iii) lim (XnYn) = XY
(all limits n going to infinity)

Thanks
Theorem: $\lim (x + y) = \lim x + \lim y$

since $\lim_{n \to \infty}x_n = X$ and $\lim_{n \to \infty}y_n = Y$

we have, $\lim_{n \to \infty} (x_n + y_n) = \lim_{n \to \infty}x_n + \lim_{n \to \infty}y_n = X + Y$

now look at the limit theorems and try the others

4. Originally Posted by taypez
iii) lim (XnYn) = XY
The second one is really easy try that one yourself. This one is the hardest of all of them.

Lemma: Convergent sequences are bounded.

Proof: Let $\{ x_n \}\to x$ be a convergent sequence then for $\epsilon = 1$ we have $|x_n-x|< 1$ for $x\geq N\in \mathbb{N}^+$. Thus, $|x_n|\leq |x_n-x|+|x| <1+|x|$. This means $x_N,x_{N+1},...$ are all bounded by $|x|+1$. So let $A = \max\left\{ |x_1|,|x_2|,...,|x_N|,|x|+1 \right\}$ then it means $|x_n|\leq A$ for all $n=1,2,3,...$. Q.E.D.

Now we can prove your problem.

We know that the sequence $y_n$ is bounded by some positive number $A$.
This means,
$|x_n-x| < \frac{\epsilon}{A+|x|} \mbox{ for }n>N_1$
$|y_n-y| < \frac{\epsilon}{A+|x|} \mbox{ for }n>N_2$.
Thus,
$|x_n-x|<\frac{\epsilon}{A+|x|} \mbox{ and }|y_n-y|< \frac{\epsilon}{A+|x|} \mbox{ for }n>N \mbox{ where }N=\max \{ N_1,N_2 \}$.

This means,
$|x_ny_n-xy| \leq |x_ny_n-xy_n|+|xy_n-xy|$ $= |y_n||x_n-x|+|x||y_n-y| < A\cdot \frac{\epsilon}{A+|x|} + |x|\cdot \frac{\epsilon}{A+|x|} = \epsilon \mbox{ for }n>N$.
Q.E.D.

Doesn't that just show that
lim (Xn + Yn) = lim Xn + lim Yn?
How do I show that lim Xn goes to X?
Also, since your proof is based on convergence, will I need to prove convergence?

6. Originally Posted by taypez
How do I show that lim Xn goes to X?
Also, since your proof is based on convergence, will I need to prove convergence?
The problem says you can assume it. Meaning it says "If xn and yn are convergent sequences ... " this means we are accepting that these sequences converge no need to prove that.

Doesn't that just show that
lim (Xn + Yn) = lim Xn + lim Yn?
How do I show that lim Xn goes to X?
NEVERMIND. DUH!

Also, since your proof is based on convergence, will I need to prove convergence.

8. Originally Posted by taypez
lim (Xn + Yn) = lim Xn + lim Yn?
How do I show that lim Xn goes to X?
Also, since your proof is based on convergence, will I need to prove convergence?
Originally Posted by taypez
... If Xn goes to X and Yn goes to Y as n goes to infinity...
..

9. Thanks!