Find \lim_{(x,y) \rightarrow (2,-3)} \frac{-5x^2-2y^2-1}{x^2+y^2-5}


Find {eq}\displaystyle \lim_{(x,y) \rightarrow (2,-3)} \frac{-5x^2-2y^2-1}{x^2+y^2-5} {/eq}

Answer and Explanation:

The function {eq}\displaystyle \frac{-5x^2-2y^2-1}{x^2+y^2-5} {/eq} has continuity at {eq}(x,y) =(2,-3) {/eq} so we directly substitute {eq}x=2 {/eq} and {eq}y=-3 {/eq}:

{eq}\begin{align*} \displaystyle \lim_{(x,y) \rightarrow (2,-3)} \frac{-5x^2-2y^2-1}{x^2+y^2-5} & = \frac{-5(2)^2-2(-3)^2-1}{(2)^2+(-3)^2-5} \;\;\; \left[ \mathrm{ Direct \ Substitution }\right]\\ & = \frac{-20-18-1}{4+9-5} \\ \implies \displaystyle \lim_{(x,y) \rightarrow (2,-3)} \frac{-5x^2-2y^2-1}{x^2+y^2-5}& = \frac{-39}{8} \\ \end{align*} {/eq}

Learn more about this topic:

How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 7 / Lesson 4

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