# Why is the limit 2025? a simple plug and chug limit! gone wrong!

1. Oct 19, 2005

### mr_coffee

Hello everyone i have the following prlbem:
lim (x,y)->(5,-2) (x^5+4x^3y-5xy^2);

So i let y be fixed at 0 and let x equal 5, and i got 5^5 which is 3125, if i let 0, and y be -2 i get 0. Doesn't this mean the limit doesn't exist? but it says its 2025, ?

2. Oct 19, 2005

### whozum

Just plug in 5 for each x value and -2 for each y value, you get 2025. You were doin something completely different.

3. Oct 19, 2005

### mr_coffee

ohhh i did that before and i messed up on multipcation thank u

4. Oct 20, 2005

### AKG

What you're doing in your first post is like saying that if you have:

$$\lim _{x \to 7} x^2$$

and you plug in x = 3 to get 9 and plug in x = 10 to get 100, then the limit doesn't exist.