MHB How do you calculate limits in two variables?

[mathmari]

Hey!

I have to calculate the following limits, if they exists.

1. $$\lim_{(x, y) \rightarrow (0, 0)} (x^2+y^2+3)$$
2. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{xy}{x^2+y^2+2}$$
3. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{e^xy}{x+1}$$
4. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{\cos x-1-x^2/2}{x^4+y^4}$$
5. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{(x-y)^2}{x^2+y^2}$$

I have done the following:

1. $x^2+y^2+3$ is continuous for each $(x, y) \in \mathbb{R}^2$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} (x^2+y^2+3)=0^2+0^2+3=3$
2. $\frac{xy}{x^2+y^2+2}$ is continuous for each $(x, y) \in \mathbb{R}^2$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} \frac{xy}{x^2+y^2+2}=\frac{0 \cdot 0}{0^2+0^2+2}=2$
3. $\frac{e^xy}{x+1}$ is continuous for $(x, y) \in \mathbb{R}^2 \setminus \{(-1, y)\}$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} \frac{e^xy}{x+1}=\frac{e^0 0}{0+1}=0$

Is ths correct?? (Wondering) Could I improve something at the formulation?? (Wondering)

What could we do at the limits $4-5$ ?? (Wondering)

 

[Prove It]

Hey!

I have to calculate the following limits, if they exists.

1. $$\lim_{(x, y) \rightarrow (0, 0)} (x^2+y^2+3)$$
2. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{xy}{x^2+y^2+2}$$
3. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{e^xy}{x+1}$$
4. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{\cos x-1-x^2/2}{x^4+y^4}$$
5. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{(x-y)^2}{x^2+y^2}$$

I have done the following:

1. $x^2+y^2+3$ is continuous for each $(x, y) \in \mathbb{R}^2$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} (x^2+y^2+3)=0^2+0^2+3=3$
2. $\frac{xy}{x^2+y^2+2}$ is continuous for each $(x, y) \in \mathbb{R}^2$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} \frac{xy}{x^2+y^2+2}=\frac{0 \cdot 0}{0^2+0^2+2}=2$
3. $\frac{e^xy}{x+1}$ is continuous for $(x, y) \in \mathbb{R}^2 \setminus \{(-1, y)\}$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} \frac{e^xy}{x+1}=\frac{e^0 0}{0+1}=0$

Is ths correct?? (Wondering) Could I improve something at the formulation?? (Wondering)

What could we do at the limits $4-5$ ?? (Wondering)

For 2. last time I checked, 0/2 = 0, not 2...

 

[Prove It]

Hey!

I have to calculate the following limits, if they exists.

1. $$\lim_{(x, y) \rightarrow (0, 0)} (x^2+y^2+3)$$
2. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{xy}{x^2+y^2+2}$$
3. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{e^xy}{x+1}$$
4. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{\cos x-1-x^2/2}{x^4+y^4}$$
5. $$\lim_{(x, y) \rightarrow (0, 0)} \frac{(x-y)^2}{x^2+y^2}$$

I have done the following:

1. $x^2+y^2+3$ is continuous for each $(x, y) \in \mathbb{R}^2$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} (x^2+y^2+3)=0^2+0^2+3=3$
2. $\frac{xy}{x^2+y^2+2}$ is continuous for each $(x, y) \in \mathbb{R}^2$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} \frac{xy}{x^2+y^2+2}=\frac{0 \cdot 0}{0^2+0^2+2}=2$
3. $\frac{e^xy}{x+1}$ is continuous for $(x, y) \in \mathbb{R}^2 \setminus \{(-1, y)\}$, so we have that $\lim_{(x, y) \rightarrow (0, 0)} \frac{e^xy}{x+1}=\frac{e^0 0}{0+1}=0$

Is ths correct?? (Wondering) Could I improve something at the formulation?? (Wondering)

What could we do at the limits $4-5$ ?? (Wondering)

5. is just screaming to be converted to polars...

$\displaystyle \begin{align*} \lim_{(x,y) \to (0,0)} \frac{(x-y)^2}{x^2+y^2} &= \lim_{r\to 0}\frac{\left[r\cos{(\theta )} - r\sin{(\theta )} \right] ^2}{r^2} \\ &= \lim_{r \to 0} \frac{r^2 \left[ \cos{(\theta )} - \sin{(\theta )} \right] ^2}{r^2} \\ &= \lim_{r \to 0} \left[ \cos{(\theta )} - \sin{(\theta )} \right] ^2 \\ &= \left[ \cos{(\theta )} - \sin{(\theta )} \right] ^2 \end{align*}$

Since this value changes according to the angle approached on, that means different paths yield different values, and thus the limit does not exist.

 

[mathmari]

For the first three one could we formulate it as followed?? (Wondering)

1. $(x, y) \rightarrow (0, 0)$
   $x \rightarrow 0$ and $y \rightarrow 0$
   So, $x^2+y^2+3 \rightarrow 0^2+0^2+3=3$
   
2. $(x, y) \rightarrow (0, 0)$
   $x \rightarrow 0$ and $y \rightarrow 0$
   So, $\frac{xy}{x^2+y^2+2} \rightarrow \frac{0 \cdot 0}{0^2+0^2+2}=0$
   
3. $(x, y) \rightarrow (0, 0)$
   $x \rightarrow 0$ and $y \rightarrow 0$
   So, $\frac{e^xy}{x+1} \rightarrow \frac{e^0 \cdot 0}{0+1}=0$

What could we do at the limit $4.$ to check if it exists?? (Wondering)

 

[I like Serena]

For the first three one could we formulate it as followed?? (Wondering)

1. $(x, y) \rightarrow (0, 0)$
   $x \rightarrow 0$ and $y \rightarrow 0$
   So, $x^2+y^2+3 \rightarrow 0^2+0^2+3=3$
   
2. $(x, y) \rightarrow (0, 0)$
   $x \rightarrow 0$ and $y \rightarrow 0$
   So, $\frac{xy}{x^2+y^2+2} \rightarrow \frac{0 \cdot 0}{0^2+0^2+2}=0$
   
3. $(x, y) \rightarrow (0, 0)$
   $x \rightarrow 0$ and $y \rightarrow 0$
   So, $\frac{e^xy}{x+1} \rightarrow \frac{e^0 \cdot 0}{0+1}=0$

What could we do at the limit $4.$ to check if it exists?? (Wondering)

Hey mathmari! (Mmm)

What you had in the OP seemed fine to me.
This also looks fine. (Nod)

For 4 and 5, I'd check what happens if you approach the origin along different lines, say $y=0$ and $x=0$. (Thinking)

 

[mathmari]

Ok... Thanks a lot! (Mmm)