Best solutions to - "How can I prove thatlim(x,y)→(0,0)|x|32y2x4+y2→0"

Lesnaoq73

Answered question

2022-03-27

How can I prove that

lim_{(x, y) \to (0, 0)} \frac{{| x |}^{\frac{3}{2}} y^{2}}{x^{4} + y^{2}} \to 0

Ronald Martinez

Beginner2022-03-28Added 7 answers

I assume you want to show that

lim_{(x, y) \to (0, 0)} \frac{{| x |}^{\frac{3}{2}} y^{2}}{x^{4} + y^{2}} \to 0

Although sometimes it is not the nicest way, switching to polar coordinates is a general approach that will solve these kinds of limits. Let

x = r \cos θ, y = r \sin θ

. Then your limit is

lim_{(x, y) \to (0, 0)} \frac{{| x |}^{\frac{3}{2}} y^{2}}{x^{4} + y^{2}} = lim_{r \to 0} \frac{r^{\frac{3}{2}} {| \cos θ |}^{\frac{3}{2}} r^{2} \sin^{2} θ}{r^{2} (r^{2} \cos^{4} θ + \sin^{2} θ)} = lim_{r \to 0} r^{\frac{3}{2}} \frac{{| \cos θ |}^{\frac{3}{2}}}{(r^{2} \frac{\cos^{4} θ}{\sin^{2} θ} + 1)}

Now, since

0 \leq \frac{{| \cos θ |}^{\frac{3}{2}}}{(r^{2} \frac{\cos^{4} θ}{\sin^{2} θ} + 1)} \leq {| \cos θ |}^{\frac{3}{2}} \leq 1

we can apply the squeeze theorem, and since

lim_{r \to 0} r^{\frac{3}{2}} = 0

, we see that the original limit is 0.
Hope that helps.

Jeffrey Jordon

Expert2022-08-30Added 2605 answers

Answer is given below (on video)

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