Problem 57 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{(x, y, z) \rightarrow(1,1,1)} \frac{x^{2}+x y-x z-y z}{x-z}$$

Step-by-Step Solution

Answer

Answer: The limit of the given expression as (x, y, z) approaches (1,1,1) is 2.

Step 1: Simplify the given expression

We can begin by factoring out a common term from the numerator of the expression: $$\frac{x^2+xy-xz-yz}{x-z} = \frac{x(x+y)-z(x+y)}{x-z}$$ Now, we can factor out \((x+y)\): $$\frac{(x-y)(x+y)}{x-z}$$

Step 2: Check for continuity at the point (1,1,1)

Now, we need to make sure that the function is continuous at the point (1,1,1). By looking at the simplified expression, we can see that it is continuous when the denominator is not equal to zero, since a zero denominator would cause the expression to be undefined. To check for continuity at the point (1,1,1), substitute the point's coordinates into the denominator and verify that it is not equal to zero: $$x-z = 1-1 = 0$$ The denominator is equal to zero at this point, which means that, before moving forward, we must remove the common factor (if any) between the numerator and the denominator that is causing this get zero at the given point.

Step 3: Removing common factor in the denominator

Observe that the factor \((x-y)\) in the numerator has a value equal to the denominator \((x-z)\) at the point (1,1,1), so we can substitute (x-y) for (x-z) in our expression: $$\frac{(x-y)(x+y)}{x-y}$$ Now, we can cancel out the common factor \((x-y)\): $$x+y$$

Step 4: Evaluate the limit

As we have a simplified expression that is continuous at the point (1,1,1), we can evaluate the limit by directly substituting the point's coordinates into the expression: $$\lim _{(x, y, z) \rightarrow(1,1,1)} x+y = 1+1 = 2$$ Thus, the limit of the given expression as \((x, y, z)\) approaches \((1,1,1)\) is 2.