Question: "Determine which of the following limits exist, and..."

rastafarral6

Answered question

2021-11-21

Determine which of the following limits exist, and find the limits which do exists.
a)

lim_{(x, y) \to (0, 0)} \frac{x^{3} + y^{3}}{x^{2} + y^{2}}

lim_{(x, y) \to (0, 0)} \frac{x^{2} + 4 x y^{2} + 4 y^{4}}{x^{2} + 4 y^{4}}

Ourst1977

Beginner2021-11-22Added 21 answers

Consider the provided question,
a) Given that,

lim_{(x, y) \to (0, 0)} \frac{x^{3} + y^{3}}{x^{2} + y^{2}}

Put y=mx

lim_{(x, y) \to (0, 0)} \frac{x^{3} + y^{3}}{x^{2} + y^{2}} = lim_{(x, m x) \to (0, 0)} \frac{3^{3} + {(m x)}^{3}}{x^{2} + {(m x)}^{2}}

= lim_{x \to 0} \frac{x^{3} (1 + m^{3})}{x^{2} (1 + m^{2})}

= lim_{x \to 0} \frac{x (1 + m^{3})}{(1 + m^{2})} = 0

(Limits is 0 for all values of m)
Hence, Limit exist and

lim_{(x, y) \to (0, 0)} \frac{x^{3} + y^{3}}{x^{2} + y^{2}} = 0

Coldst

Beginner2021-11-23Added 18 answers

b) Given that,

lim_{(x, y) \to (0, 0)} \frac{x^{2} + 4 x y^{2} + 4 y^{4}}{x^{2} + 4 y^{4}}

Put y=mx

lim_{(x, y) \to (0, 0)} \frac{x^{2} + 4 x y^{2} + 4 y^{4}}{x^{2} + 4 y^{4}} = lim_{(x, m x) \to (0, 0)} \frac{x^{2} + 4 x {(m x)}^{2} + 4 {(m x)}^{4}}{x^{2} + 4 {(m x)}^{4}}

= lim_{x \to 0} \frac{x^{2} + 4 m^{2} x^{3} + 4 m^{4} x^{4}}{x^{2} + 4 m^{4} x^{4}}

= lim_{x \to 0} \frac{x^{2} (1 + 4 m^{2} x + 4 m^{4} x^{2})}{x^{2} (1 + 4 m^{4} x^{2})}

= lim_{x \to 0} \frac{(1 + 4 m^{2} x + 4 m^{4} x^{2})}{(1 + 4 m^{4} x^{2})}

= lim_{x \to 0} \frac{{(1 + 2 m^{2} x)}^{2}}{(1 + 4 m^{4} x^{2})}

= \frac{{(1 + 0)}^{2}}{1 + 0}

= 1

Thus, here limit exist and

lim_{(x, y) \to (0, 0)} \frac{x^{2} + 4 x y^{2} + 4 y^{4}}{x^{2} + 4 y^{4}} = 1

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