Evaluate lim (x,y)→(1,0) (x - 1)² ln x / ((x - 1)² + y²) and lim (x,y)→(0,0) 2x²y / (x⁴ + y²).

Steps to Solve

1. Evaluate the first limit
   We want to find:
   $$ \lim_{(x,y) \to (1,0)} \frac{(x - 1)^2 \ln x}{(x - 1)^2 + y^2} $$
   Substituting ( x = 1 ) gives ( (1 - 1)^2 \ln 1 = 0 ), and for ( y = 0 ), we have ( (1 - 1)^2 + 0^2 = 0 ) in the denominator. Thus, we have an indeterminate form ( \frac{0}{0} ).

2. Apply L'Hôpital's Rule for the first limit
   Since we have an indeterminate form, we can apply L'Hôpital's Rule. Differentiate the numerator and the denominator with respect to ( x ) while treating ( y ) as constant:

• Derivative of the numerator:
  $$ \frac{d}{dx}[(x-1)^2 \ln x] = 2(x-1) \ln x + (x-1) $$
• Derivative of the denominator:
  $$ \frac{d}{dx}[(x-1)^2 + y^2] = 2(x-1) $$

Thus our limit becomes:
$$ \lim_{(x,y) \to (1,0)} \frac{2(x-1) \ln x + (x-1)}{2(x-1)} $$

3. Simplify and substitute for the first limit
   We can simplify:
   $$ = \lim_{(x,y) \to (1,0)} \frac{\ln x + 1}{2} $$
   Substituting ( x = 1 ):
   $$ = \frac{\ln 1 + 1}{2} = \frac{0 + 1}{2} = \frac{1}{2} $$

4. Evaluate the second limit
   Next limit to evaluate:
   $$ \lim_{(x,y) \to (0,0)} \frac{2x^2 y}{x^4 + y^2} $$
   Substituting ( x = 0 ) and ( y = 0 ) gives ( \frac{0}{0} ).

5. Convert to polar coordinates for the second limit
   In polar coordinates, set ( x = r \cos \theta ) and ( y = r \sin \theta ), and as ( r \to 0 ):
   $$ = \lim_{r \to 0} \frac{2(r \cos \theta)^2 (r \sin \theta)}{(r \cos \theta)^4 + (r \sin \theta)^2} $$
   This simplifies to:
   $$ = \lim_{r \to 0} \frac{2r^3 \cos^2 \theta \sin \theta}{r^4 \cos^4 \theta + r^2 \sin^2 \theta} = \lim_{r \to 0} \frac{2r \cos^2 \theta \sin \theta}{r^2 \cos^4 \theta + \sin^2 \theta} $$

6. Evaluate the limit for the second limit
   As ( r \to 0 ), the limit becomes:
   $$ = \frac{0}{\sin^2 \theta} = 0 $$

Thus, the final limits are:

1. The first limit is ( \frac{1}{2} )
2. The second limit is ( 0 )