Question

1. analizar la existencia de los siguientes límites. si existen dar su valor.

(a) \\(\lim_{(x,y) \to (-1,1)} \frac{(x+1)\sin^2(y-1)}{(x+1)^2 + (y-1)^2}\\);

(b) \\(\lim_{(x,y) \to (1,0)} \frac{\sqrt{|y|}(x-1)(x+4)}{(x-1)^2 + y}\\).

Explanation:

🆕 New Concept Discovered: Multivariable Limits
Analyzing limits in two variables using substitution and path testing.

Step 1: Simplify part (a) using variable substitution

To analyze the limit

$$ \lim_{(x,y) \to (-1,1)} \frac{(x+1)\sin^2(y-1)}{(x+1)^2 + (y-1)^2} $$

we can simplify the expression by shifting the limit point to the origin \( (0,0) \). Let:

$$ u = x + 1 \quad \text{and} \quad v = y - 1 $$

As \( (x,y) \to (-1,1) \), we have \( (u,v) \to (0,0) \). Substituting these into the limit gives:

$$ \lim_{(u,v) \to (0,0)} \frac{u \sin^2(v)}{u^2 + v^2} $$

Step 2: Analyze part (a) using polar coordinates

We can convert the limit to polar coordinates where \( u = r\cos(\theta) \) and \( v = r\sin(\theta) \), with \( r \to 0 \):

$$ \lim_{r \to 0} \frac{r\cos(\theta) \sin^2(r\sin(\theta))}{r^2} $$

Using the fundamental trigonometric limit \( \sin(w) \approx w \) as \( w \to 0 \), we can approximate \( \sin^2(r\sin(\theta)) \approx r^2\sin^2(\theta) \) for small \( r \):

$$ \lim_{r \to 0} \frac{r\cos(\theta) \cdot r^2\sin^2(\theta)}{r^2} = \lim_{r \to 0} r\cos(\theta)\sin^2(\theta) $$

Since \( |\cos(\theta)\sin^2(\theta)| \le 1 \) is bounded, taking the limit as \( r \to 0 \) yields:

$$ 0 $$

Thus, the limit for part (a) exists and its value is \( 0 \).

Step 3: Simplify part (b) using variable substitution

For the second limit:

$$ \lim_{(x,y) \to (1,0)} \frac{\sqrt{|y|}(x-1)(x+4)}{(x-1)^2 + y} $$

Let \( u = x - 1 \), which means \( x = u + 1 \). As \( (x,y) \to (1,0) \), we have \( (u,y) \to (0,0) \). Substituting these into the expression gives:

$$ \lim_{(u,y) \to (0,0)} \frac{\sqrt{|y|} \cdot u \cdot (u+5)}{u^2 + y} $$

Step 4: Test paths for part (b) to check existence

To see if this limit exists, we can approach the origin \( (0,0) \) along different paths.

Path 1: Approach along the curve \( y = m u^2 \) (where \( m > 0 \)):

$$ \lim_{u \to 0} \frac{\sqrt{|m u^2|} \cdot u \cdot (u+5)}{u^2 + m u^2} = \lim_{u \to 0} \frac{\sqrt{m}|u| \cdot u(u+5)}{u^2(1+m)} $$

If we approach from the positive side \( u \to 0^+ \), then \( |u| = u \):

$$ \lim_{u \to 0^+} \frac{\sqrt{m} u^2 (u+5)}{u^2(1+m)} = \lim_{u \to 0^+} \frac{\sqrt{m}(u+5)}{1+m} = \frac{5\sqrt{m}}{1+m} $$

Since this value depends directly on the choice of \( m \) (the path chosen), approaching along different parabolic paths yields different limiting values.

Because the limit value is not unique, the limit in part (b) does not exist.

Answer:

(a) The limit exists and its value is \( 0 \).

(b) The limit does not exist.[LLM SSE On Failure]