limxy→01x+ylnx^2+e^y

To evaluate this limit, we can try to approach the point (0,1) along different paths and see if the limit is the same along all paths. If it is, then we have found the limit. If not, then the limit does not exist.

Let's try approaching (0,1) along the x-axis, y-axis, and the line y=x.

Approaching along the x-axis:

Let y=0. Then we have lim(x→0)(x+0)ln(x^2+e^0) = lim(x→0)xln(x^2+1)

As x approaches 0, ln(x^2+1) approaches ln(1) = 0, so we can use L'Hopital's rule to evaluate the limit:

lim(x→0)x*ln(x^2+1) = lim(x→0)(ln(x^2+1))/(1/x)

= lim(x→0)(2x)/(2x/(x^2+1)) = lim(x→0)(x^2+1)

= 1

So the limit along the x-axis is 0*ln(1) = 0.

Approaching along the y-axis:

Let x=0. Then we have lim(y→1)(0+y)ln(0^2+e^y) = lim(y→1)yln(e^y)

= lim(y→1)y*y = 1

So the limit along the y-axis is 1*ln(1) = 0.

Approaching along the line y=x:

Let y=x. Then we have lim(x→0)(x+x)ln(x^2+e^x) = lim(x→0)2xln(x^2+e^x)

As x approaches 0, we have ln(x^2+e^x) approaches ln(e^x) = x, so:

lim(x→0)2xln(x^2+e^x) = lim(x→0)2xx

= 0

So the limit along the line y=x is 0*ln(1) = 0.

Since the limit along each path is different, the limit as (x,y) approaches (0,1) does not exist