Does the Limit of the Function Approach Zero as (x, y) Tends to (0,0)?

In summary: Sorry for the mistake earlier!In summary, the limit of $\displaystyle \frac{x^2 - y^2}{\sqrt{x^2 + y^2}}$ as $(x, y)$ approaches $(0, 0)$ is 0. This can be shown by converting to polar coordinates and evaluating the limit $\displaystyle \lim_{r \to 0} \, r \cos{(2\theta)}$. As the value of this limit does not change depending on the path taken, the limit is 0.
  • #1
tmt1
234
0
I need to find

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

If I plug in zero, I get an indeterminate form. How do I resolve the indeterminate form?
 
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  • #2
Hi tmt,

First start with

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

Can you find this limit?
 
  • #3
tmt said:
I need to find

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

If I plug in zero, I get an indeterminate form. How do I resolve the indeterminate form?

I would convert to polars. With $\displaystyle \begin{align*} x = r\cos{ \left( \theta \right) } \end{align*}$ and $\displaystyle \begin{align*} y = r\sin{ \left( \theta \right) } \end{align*}$ this limit is

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

As the value of this limit does not change depending on the path you take (so which angle you approach the origin from) that means the limit is 0.
 
Last edited:
  • #4
Prove It said:
I would convert to polars. With $\displaystyle \begin{align*} x = r\cos{ \left( \theta \right) } \end{align*}$ and $\displaystyle \begin{align*} y = r\sin{ \left( \theta \right) } \end{align*}$ this limit is

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

As the value of this limit changes depending on the path you take (so which angle you approach the origin from) that means the limit does not exist.

In the first step, the denominator should be $$\sqrt{[r\cos(\theta)]^2 + [r\sin(\theta)]^2}$$ so instead it comes down to computing $\lim\limits_{r\to 0} r\cos(2\theta)$.
 
  • #5
Euge said:
In the first step, the denominator should be $$\sqrt{[r\cos(\theta)]^2 + [r\sin(\theta)]^2}$$ so instead it comes down to computing $\lim\limits_{r\to 0} r\cos(2\theta)$.

Oh whoops :P OK instead I showed that the limit is 0 hahaha. Will edit my post now (y)
 
  • #6
Euge said:
In the first step, the denominator should be $$\sqrt{[r\cos(\theta)]^2 + [r\sin(\theta)]^2}$$ so instead it comes down to computing $\lim\limits_{r\to 0} r\cos(2\theta)$.

So it evaluates to 0?
 
  • #7
tmt said:
So it evaluates to 0?

Yes, that's correct.
 

FAQ: Does the Limit of the Function Approach Zero as (x, y) Tends to (0,0)?

What is the definition of a limit for a 2 variable function?

A limit for a 2 variable function is the value that a function approaches as the input variables get closer and closer to a specific point. This point is usually denoted by (x,y) and is referred to as the limit point.

How is the limit of a 2 variable function calculated?

The limit of a 2 variable function is calculated by taking the limit of the function as both input variables approach the limit point simultaneously. This can be done by evaluating the function at different points that are closer and closer to the limit point, and observing the trend of the values.

What does it mean if the limit of a 2 variable function does not exist?

If the limit of a 2 variable function does not exist, it means that the function does not approach a single value as the input variables get closer to the limit point. This could be due to the function being undefined at the limit point, or the values of the function oscillating between different values as the input variables approach the limit point.

Can a 2 variable function have a different limit at different points?

Yes, a 2 variable function can have a different limit at different points. This is because the behavior of the function may vary depending on the direction in which the input variables approach the limit point. This is known as directional or one-sided limits.

What are some real-life applications of limits for 2 variable functions?

Limits for 2 variable functions have many real-life applications, such as in economics, engineering, and physics. For example, they can be used to determine the maximum profit for a company based on different levels of production, or to analyze the stability of a structure under varying loads and conditions. They are also used in calculating derivatives and integrals, which are essential in many scientific fields.

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