f(x,y) = 3x + 3y
Show that for any epsilon in:
|2x + 3y - 5| < epsilon
There is a delta value:
sqrt((x - 1)^2 + (y - 1)^2) < delta
Solve for delta in terms of epsilon
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I understand that this is a proof that the limit of f(x,y) as (x,y) -> (1,1) is 5.
But I'm stuck solving for delta in terms of epsilon.
thanks!
Sorry, f(x,y) = 2x+3y, not 3x+3y
To solve for delta in terms of epsilon, we need to rewrite the given inequality |2x + 3y - 5| < epsilon in terms of the expression involving delta.
Let's begin by simplifying the left-hand side of the inequality |2x + 3y - 5|:
|2x + 3y - 5| = |(2x - 2) + (3y - 3)| = |2(x - 1) + 3(y - 1)|
Now, let's recall the triangle inequality, which states that for any real numbers a and b:
|a + b| ≤ |a| + |b|
Applying this to our expression, we have:
|2(x - 1) + 3(y - 1)| ≤ |2(x - 1)| + |3(y - 1)|
Since x and y are variables, let's denote them as x = 1 + Δx and y = 1 + Δy, where Δx and Δy are small changes in x and y, respectively.
Substituting these values into the inequality, we get:
|2((1 + Δx) - 1)| + |3((1 + Δy) - 1)| ≤ |2Δx| + |3Δy|
Simplifying further:
|2Δx| + |3Δy|
Now, we need to relate this expression to the given expression involving delta, which is:
sqrt((x - 1)^2 + (y - 1)^2) < delta
Substituting the values of x and y as x = 1 + Δx and y = 1 + Δy, we have:
sqrt(((1 + Δx) - 1)^2 + ((1 + Δy) - 1)^2) < delta
Simplifying this expression:
sqrt(Δx^2 + Δy^2) < delta
Hence, we can see that delta in terms of epsilon is equivalent to sqrt(Δx^2 + Δy^2) < delta is the same as |2Δx| + |3Δy| < epsilon.
Therefore, we can conclude that delta = epsilon is the solution for delta in terms of epsilon.