(1 point) Find the limit, if it exists, or type DNE if it does not exist. A. lim(x,y)→(0,0)(x+5y)2x2+25y2=lim(x,y)→(0,0)(x+5y)2x2+25y2= B. lim(x,y)→(0,0)2x3+7y3x2+y2=lim(x,y)→(0,0)2x3+7y3x2+y2= (Hint for B: use polar coordinates, that is x=rcos(θ),y=rsin(θ)x=rcos⁡(θ),y=rsin⁡(θ) )

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adamu4mohammed adamu4mohammed · Answer 1 · 2020-03-25T07:31:40+01:00

Answer:

A. DNE. The limit does not exist.

B. The limit is 0

Step-by-step explanation:

A. lim(x,y)→(0,0)[(x + 5y)/(2x² + 25y²)]

Along the x axis, y = 0, the limit becomes:

lim(x)→(0)[x/2x²]

= lim(x)→(0)[1/2x] = infinity

Along the y axis, x = 0, the limit becomes

lim(y)→(0)[5y/25y²]

= lim(y)→(0)[1/5y] = infinity

DNE

B. lim(x,y)→(0,0)[(2x³ + 7y³)/(x² + y²)

Putting x = rcosθ, and y = rsinθ, the limit becomes

lim(r)→(0)[(2r³cos³θ + 7r³sin³θ)/(r²cos²θ + r²sin²θ)]

Because cos²θ + sin²θ = 1, the limit becomes:

= lim(r)→(0)[r³(2cos³θ + 7sin³θ)/r²]

= lim(r)→(0)[r(2cos³θ + 7sin³θ)]

= 0

The limit is 0