How do you find the limit as (x,y) approaches (0,0) of #(x+y^2) / (2x+y)#?

Answer 1

#lim_((x,y)->(0,0)) (x+ y^2)/(2x + y)#

#ln [lim_((x,y)->(0,0)) (x+ y^2)/(2x + y)]#

#lim_((x,y)->(0,0)) ln[(x+ y^2)/(2x + y)]#

#lim_((x,y)->(0,0)) ln(x+ y^2) - ln(2x + y)#

#lim_((x,y)->(0,0)) ln(x+ y^2) - lim_((x,y)->(0,0)) ln(2x + y)#

#= e^(lim_((x,y)->(0,0)) ln(x+ y^2) - lim_((x,y)->(0,0)) ln(2x + y))#

#= e^(oo - oo)#

You cannot do #oo - oo#.

There is no way you can rewrite this without it be undefined. This is the graph:

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Answer 2

To find the limit as (x,y) approaches (0,0) of (x+y^2) / (2x+y), we can use the limit definition. First, we substitute the values of x and y into the expression:

lim(x,y)→(0,0) (x+y^2) / (2x+y)

Next, we can try approaching the limit along different paths. Let's consider approaching along the x-axis (y=0):

lim(x,y)→(0,0) (x+0^2) / (2x+0) lim(x,y)→(0,0) x / (2x) lim(x,y)→(0,0) 1/2

Now, let's approach along the y-axis (x=0):

lim(x,y)→(0,0) (0+y^2) / (0+y) lim(x,y)→(0,0) y^2 / y lim(x,y)→(0,0) y

Since the limit approaches different values along different paths, the limit does not exist as (x,y) approaches (0,0) for the given expression.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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