Exploring Limits: Calculating Limits in Two Variables

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In summary, there is a conversation about calculating limits at the origin for various equations. The expert suggests using polar coordinates for the fifth equation and checking different approaches for the fourth and fifth equations to see if the limits exist.
  • #1
mathmari
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Hey! :eek:

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)
 
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  • #2
mathmari said:
Hey! :eek:

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...
 
  • #3
mathmari said:
Hey! :eek:

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.
 
  • #4
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)
 
  • #5
mathmari said:
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)
 
  • #6
Ok... Thanks a lot! (Mmm)
 

FAQ: Exploring Limits: Calculating Limits in Two Variables

What is a limit?

A limit is a mathematical concept that represents the value that a function approaches as the input approaches a certain value. It is used to describe the behavior of a function near a specific point.

How do you calculate a limit?

To calculate a limit, you can use various methods, such as substitution, factoring, and rationalization. You can also use the limit laws, which state that the limit of a sum, difference, product, or quotient of two functions is equal to the sum, difference, product, or quotient of their limits.

Why are limits important?

Limits are important because they allow us to understand the behavior of a function near a specific point. They are also used in calculus to define derivatives and integrals, which are fundamental concepts in mathematics and are essential in many real-world applications.

What are the common types of limits?

The common types of limits include one-sided limits, where the input approaches the specific point from either the left or the right side, and infinite limits, where the input approaches positive or negative infinity. There are also limits at infinity, where the input approaches infinity, and limits of trigonometric functions, logarithmic functions, and exponential functions.

How can limits be used to solve real-world problems?

Limits can be used to solve real-world problems by helping us understand the behavior of a system near a specific point. For example, they can be used to determine the maximum or minimum value of a function, which can be useful in optimizing processes in engineering and economics. They can also be used to model the growth and decay of populations, the spread of diseases, and the change in temperature over time.

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