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Find the limit of f as or show that the limit does not exist. Consider converting the function to polar coordinates to make finding the limit easier. f(x,y)

Sagot :

MrRoyal

Answer:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Step-by-step explanation:

Given

[tex]f(x,y) = \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Required

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex]

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex] becomes

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Multiply by 1

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}\cdot 1[/tex]

Express 1 as

[tex]\frac{y^2}{y^2} = 1[/tex]

So, the expression becomes:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} \cdot \frac{y^2}{y^2}[/tex]

Rewrite as:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} \cdot \frac{\sin^2y}{y^2}[/tex]

In limits:

[tex]\lim_{(x,y) \to (0,0)} \frac{\sin^2y}{y^2} \to 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} *1[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2}[/tex]

Convert to polar coordinates; such that:

[tex]x = r\cos\theta;\ \ y = r\sin\theta;[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{(r\cos\theta)^2 (r\sin\theta;)^2}{(r\cos\theta)^2+2(r\sin\theta;)^2}[/tex]

Expand

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2\cos^2\theta+2r^2\sin^2\theta}[/tex]

Factor out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2(\cos^2\theta+2\sin^2\theta)}[/tex]

Cancel out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

Express [tex]2\sin^2 \theta[/tex] as [tex]\sin^2\theta+\sin^2\theta[/tex]

So:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+\sin^2\theta+\sin^2\theta}[/tex]

In trigonometry:

[tex]\cos^2\theta + \sin^2\theta = 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

Evaluate the limits by substituting 0 for r

[tex]\frac{0^2 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0}{1+\sin^2\theta}[/tex]

Since the denominator is non-zero; Then, the expression becomes 0 i.e.

[tex]\frac{0}{1+\sin^2\theta} = 0[/tex]

So,

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

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