Let {x_n} and {y_n} be real valued sequences suppose x_n->0 and {y_n} is bounded. Prove that (x_n*y_n)->0

To prove that the sequence (x_n * y_n) approaches 0, given that the sequence x_n approaches 0 and y_n is bounded, we can use the concept of limit of a product and the properties of limits.

First, since x_n approaches 0, we can use the epsilon-delta definition of the limit to say that for every positive real number epsilon, there exists a positive integer N such that for all n > N:

| x_n - 0 | < epsilon/ M1 (1)

where M1 is a positive constant bound for the sequence {y_n}.

Second, since y_n is bounded, we know that there exists a positive constant M2 such that for all n:

|y_n| ≤ M2 (2)

To prove that (x_n * y_n) approaches 0, again using the epsilon-delta definition of a limit, we want to show that for every positive real number epsilon, there exists a positive integer N' such that for all n > N':

| (x_n * y_n) - 0 | < epsilon (3)

We can rewrite (3) using the triangle inequality:

| (x_n * y_n) - 0 | = | x_n * y_n | ≤ | x_n | | y_n |

Now, applying the bounds from (1) and (2), we have:

| (x_n * y_n) - 0 | ≤ | x_n | | y_n | ≤ (epsilon / M1) * M2

To make this upper bound less than epsilon, we can choose N' such that M2 * (epsilon / M1) is less than epsilon. This is possible because M1 and M2 are both positive constants and epsilon is a positive real number.

Therefore, we can conclude that for every positive real number epsilon, there exists a positive integer N' such that for all n > N', | (x_n * y_n) - 0 | < epsilon. This shows that (x_n * y_n) approaches 0 as n goes to infinity.