Jaden Easton

## Answered question

2021-01-17

Evaluate the following limits.
$\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{yz-xy-xz-{x}^{2}}{yz+xy+xz-{y}^{2}}$

### Answer & Explanation

Jozlyn

Skilled2021-01-18Added 85 answers

Evaluate:
$\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{yz-xy-xz-{x}^{2}}{yz+xy+xz-{y}^{2}}$
Simplification:
We have
$\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{yz-xy-xz-{x}^{2}}{yz+xy+xz-{y}^{2}}$
Apply the limit,
$\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{yz-xy-xz-{x}^{2}}{yz+xy+xz-{y}^{2}}=\frac{\left(1\right)\left(1\right)-\left(1\right)\left(1\right)-\left(1\right)\left(1\right)-\left(1{\right)}^{2}}{\left(1\right)\left(1\right)+\left(1\right)\left(1\right)+\left(1\right)\left(1\right)-\left(1{\right)}^{2}}$
$\frac{1-1-1-1}{1+1+1-1}$
$=\frac{-2}{2}$
$=-1$
Hence,
$\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{yz-xy-xz-{x}^{2}}{yz+xy+xz-{y}^{2}}=-1$

Jeffrey Jordon

Expert2022-04-01Added 2607 answers

Answer is given below (on video)

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