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 limi

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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)

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Khải Quang 4 years 2021-08-06T16:06:42+00:00 1 Answers 24 views 0

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  1. trungdung22
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    2021-08-06T16:08:23+00:00
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    Answer:

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

    Step-by-step explanation:

    Given

    f(x,y) = \frac{x^2 \sin^2y}{x^2+2y^2}

    Required

    \lim_{(x,y) \to (0,0)} f(x,y)

    \lim_{(x,y) \to (0,0)} f(x,y) becomes

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

    Multiply by 1

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

    Express 1 as

    \frac{y^2}{y^2} = 1

    So, the expression becomes:

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

    Rewrite as:

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

    In limits:

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

    So, we have:

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

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

    Convert to polar coordinates; such that:

    x = r\cos\theta;\ \ y = r\sin\theta;

    So, we have:

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

    Expand

    \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}

    Factor out r^2

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

    Cancel out r^2

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

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

    Express 2\sin^2 \theta as \sin^2\theta+\sin^2\theta

    So:

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

    In trigonometry:

    \cos^2\theta + \sin^2\theta = 1

    So, we have:

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

    Evaluate the limits by substituting 0 for r

    \frac{0^2 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}

    \frac{0 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}

    \frac{0}{1+\sin^2\theta}

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

    \frac{0}{1+\sin^2\theta} = 0

    So,

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

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