Question

Evaluate the following limits. $$\lim _{(x, y, z) \rightarrow(1,-1,1)} \frac{x z+5 x+y z+5 y}{x+y}$$

Short answer

$$ Answer: The limit does not exist.

Step-by-step solution

Plug in the limit point and simplify the numerator

To do this, substitute \(x=1\), \(y=-1\), \(z=1\) into the given expression: $$\frac{(1)(1)+5(1)+(-1)(1)+5(-1)}{1-1}$$

Simplify the expression

Simplify the expression by combining like terms in the numerator and denominator: $$\frac{1+5-1-5}{0}$$ Simplify the fraction: $$\frac{0}{0}$$ Since we have an indeterminate form \(\frac{0}{0}\), we will need to explore other techniques to find the limit.

Apply L'Hospital's Rule (multivariate version)

Unfortunately, L'Hospital's Rule is not directly applicable in this multivariate case. Instead, let's find a different path and use polar coordinates for solving this limit issue.

Switch to polar coordinates

Replace \((x, y, z)\) with polar coordinates \((r, \theta, \phi)\), where: $$x = 1 + r\cos{\theta}\sin{\phi}$$ $$y = -1 + r\sin{\theta}\sin{\phi}$$ $$z = 1 + r\cos{\phi}$$ As \((x, y, z) \rightarrow(1,-1,1)\), we have \((r, \theta, \phi) \rightarrow(0, \theta, \phi)\).

Substitute the polar coordinates into the expression

Replace \(x\), \(y\), and \(z\) by their polar equivalents: $$\frac{(1+r\cos\theta\sin\phi)(1+r\cos\phi)+5(1+r\cos\theta\sin\phi)+(-1+r\sin\theta\sin\phi)(1+r\cos\phi)+5(-1+r\sin\theta\sin\phi)}{(1+r\cos\theta\sin\phi)+(-1+r\sin\theta\sin\phi)}$$

Evaluate the limit as \(r \rightarrow 0\)

We can now try to find the limit as \(r \rightarrow 0\): $$\lim _{r \rightarrow 0} \frac{(1+r\cos\theta\sin\phi)(1+r\cos\phi)+5(1+r\cos\theta\sin\phi)+(-1+r\sin\theta\sin\phi)(1+r\cos\phi)+5(-1+r\sin\theta\sin\phi)}{(1+r\cos\theta\sin\phi)+(-1+r\sin\theta\sin\phi)}$$ Notice that all terms containing \(r\) in both the numerator and the denominator will go to zero as \(r\rightarrow 0\), leaving only the constants. Thus, the expression becomes: $$\frac{(1)(1)+5(1)+(-1)(1)+5(-1)}{(1)+(-1)}$$ Which is exactly the same expression as in Step 2. Unfortunately, we could not avoid the indeterminate form \(\frac{0}{0}\). Therefore, the provided limit does not exist.