Finding the K for the limit of a sequence

[Design]

Homework Statement

Let {x_k} = 3k+4/k-5; then lim k-> infinity {x_k} = 3. Given E > 0 , find an integer K so that |{x_k}-3|<E when k>K

The Attempt at a Solution

| 3k+4/k-5 - 3| < E
| 3k+4/k-5 | < E + 3

Now I don't know how to isolate the left part to a single k to find the value of k and know the value of K.

 

[JThompson]

Don't add 3 to both sides- instead, combine into a single fraction

$$\left|\frac{3k+4}{k-5}-3\right|=\left|\frac{3k+4-3(k-5)}{k-5}\right|=\left|\frac{19}{k-5}\right|<E$$

 

[Design]

I see, thanks