Math  /  Calculus

QuestionSuppose that lim(x,y)(5,2)f(x,y)=7\lim _{(x, y) \rightarrow(5,2)} f(x, y)=7. What can you say about the value of f(5,2)f(5,2) ? We can say f(5,2)=f(5,2)=\infty. We can say f(5,2)f(5,2) is an open point. We can say lim(x,y)(5,2)f(x,y)=f(5,2)=7\lim _{(x, y) \rightarrow(5,2)} f(x, y)=f(5,2)=7. We can say (x,y)(5,2)f(x,y)=f(5,2)=\underset{(x, y) \rightarrow(5,2)}{ } f(x, y)=f(5,2)=\infty. In general, nothing can be said about the value of f(5,2)f(5,2).
What if ff is continuous? We can say f(5,2)=f(5,2)=\infty. We can say f(5,2)f(5,2) is an open point. We can say (x,y)(5,2)(x, y) \rightarrow(5,2). We can say lim(x,y)(5,2)f(x,y)=f(5,2)=\lim _{(x, y) \rightarrow(5,2)} f(x, y)=f(5,2)=\infty. In general, nothing can be said about the value of f(5,2)f(5,2).

Studdy Solution

STEP 1

What is this asking? If we know a function approaches a value as its input approaches a point, can we say anything about the function's *actual* value at that point?
What if the function is continuous? Watch out! Don't assume the limit of a function at a point is the same as the function's value at that point!
That's only true if the function is **continuous** at that point.

STEP 2

1. Limit Definition
2. Continuity Definition
3. Analyze the Problem

STEP 3

Let's remember what a **limit** means!
The expression lim(x,y)(5,2)f(x,y)=7\lim_{(x, y) \rightarrow (5,2)} f(x, y) = 7 means as (x,y)(x, y) gets *really* close to (5,2)(5, 2), f(x,y)f(x, y) gets *really* close to **7**.
Notice, we're talking about what happens *near* (5,2)(5, 2), not *at* (5,2)(5, 2).

STEP 4

Now, a function is **continuous** at a point if the limit of the function as the input approaches that point is *equal* to the function's value at that point.
In math terms, f(x,y)f(x, y) is continuous at (5,2)(5, 2) if lim(x,y)(5,2)f(x,y)=f(5,2)\lim_{(x, y) \rightarrow (5,2)} f(x, y) = f(5, 2).

STEP 5

The problem tells us lim(x,y)(5,2)f(x,y)=7\lim_{(x, y) \rightarrow (5,2)} f(x, y) = 7.
This tells us *nothing* about f(5,2)f(5, 2)!
The function could be defined at (5,2)(5, 2), undefined at (5,2)(5, 2), or even be a completely different value like f(5,2)=1000f(5, 2) = 1000!

STEP 6

However, if ff is **continuous**, then by definition, the limit and the function's value at that point *must* be the same!
So, if ff is continuous at (5,2)(5, 2), then we *know* f(5,2)=7f(5, 2) = 7!

STEP 7

If we only know the limit of f(x,y)f(x, y) as (x,y)(x, y) approaches (5,2)(5, 2) is **7**, we can't say *anything* about f(5,2)f(5, 2).
But, if ff is **continuous** at (5,2)(5, 2), then we know f(5,2)=7f(5, 2) = 7.

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