Question

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)

Answer 1

`\lim_((x,y) \to (0,0)) (x^2 \sin^2y)/(x^2+2y^2) = 0`

Explanation:

Given

`f(x,y) = (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)) (x^2 \sin^2y)/(x^2+2y^2)`

Multiply by 1

`\lim_((x,y) \to (0,0)) (x^2 \sin^2y)/(x^2+2y^2)\cdot 1`

Express 1 as

`(y^2)/(y^2) = 1`

So, the expression becomes:

`\lim_((x,y) \to (0,0)) (x^2 \sin^2y)/(x^2+2y^2) \cdot (y^2)/(y^2)`

Rewrite as:

`\lim_((x,y) \to (0,0)) (x^2 y^2)/(x^2+2y^2) \cdot (\sin^2y)/(y^2)`

In limits:

`\lim_((x,y) \to (0,0)) (\sin^2y)/(y^2) \to 1`

So, we have:

`\lim_((x,y) \to (0,0)) (x^2 y^2)/(x^2+2y^2) *1`

`\lim_((x,y) \to (0,0)) (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)) ((r\cos\theta)^2 (r\sin\theta;)^2)/((r\cos\theta)^2+2(r\sin\theta;)^2)`

Expand

`\lim_((x,y) \to (0,0)) (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)) (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)) (r^2\cos^2\theta\sin^2\theta)/(\cos^2\theta+2\sin^2\theta)`

`\lim_((x,y) \to (0,0)) (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)) (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)) (r^2\cos^2\theta\sin^2\theta)/(1+\sin^2\theta)`

Evaluate the limits by substituting 0 for r

`(0^2 \cdot \cos^2\theta\sin^2\theta)/(1+\sin^2\theta)`

`(0 \cdot \cos^2\theta\sin^2\theta)/(1+\sin^2\theta)`

`(0)/(1+\sin^2\theta)`

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

`(0)/(1+\sin^2\theta) = 0`

So,

`\lim_((x,y) \to (0,0)) (x^2 \sin^2y)/(x^2+2y^2) = 0`