prove that the folllowing limits exits: and please show the work in detail becau

Question

prove that the folllowing limits exits: and please show the work in detail because I keep getting confused. Can you please tellmy the steps because I was making y=x and y=-x and I dont know why I was doing that. THANK YOU WHO EVER DOES THESE. I will throw more point. TY TY

(a) lim(x,y) to (0,0) (x^4-y^4)/(x^2+y^2)

(b)lim(x,y) to (0,0) (sin(x^2-y^2))/(x-y)

(c) lim(x,y) to (0,0) ((sin(xy))/(sqrt(x^2+y^2))

(d) THis one is tricky i thought......it asks if the limt exists....it says does lim (x,y) to(0,0) (sin(x^2+y^2))/(x^2+y^2) exist? Prove answer!!!

Explanation / Answer

a)

Limit of (x+y)/(x^2-y^2) as (x,y) approaches (0,0) =

Limit of (x+y) / [ (x-y)(x+y) ] as (x,y) approaches (0,0) =

Limit of 1 / [ (x-y) ] as (x,y) approaches (0,0) = can cancel as long as not at (0,0)

and this is undefined!

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b)

(sin(x^2-y^2))/(x-y) => (sin(x^2-y^2))/(x-y) * (x+y) / (x+y)

(sin(x^2-y^2)) * (x+y)/(x^2-y^2)

limit of (sin(x^2-y^2))/(x^2-y^2) * (x+y)

if limit (x,y) tends to (0,0) then limit of x^2 - y^2 tends to zero

so

limit sin(x^2-y^2))/(x^2-y^2) = 1

and 1 * limit (x+y) = 0 as x+y =>0

so lim(x,y) to (0,0) (sin(x^2-y^2))/(x-y) = 0

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c)

Path 1: Approach the origin along the line y = 0. Then:
lim(x,y)->(0,0) sin(xy)/(x^2+y^2)
= lim(x->0) sin(x(0))/(x^2+0^2)
= lim(x->0) 0
= 0

Path 2: Approach the origin along the line y = x. Then:
lim(x,y)->(0,0) sin(xy)/(x^2+y^2)
= lim(x->0) sin(x(x))/(x^2+x^2)
= lim(x->0) sin(x^2)/(2x^2) <--[Indeterminate form of 0/0, so use L'Hopital's Rule]
= lim(x->0) (2x)cos(x^2)/(4x)
= lim(x->0) cos(x^2)/2
= cos(0^2)/2
= 1/2

But 0 doesn't equal 1/2, so the limit doesn't exist.

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d)

lim (x,y) to(0,0) (sin(x^2+y^2))/(x^2+y^2)

lim (x,y) to(0,0) then (x^2+y^2) to 0

as lim sin(x)/x = 1 ................ x to zero

lim (x,y) to(0,0) (sin(x^2+y^2))/(x^2+y^2) = 1

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