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

[tmt1]

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?

 

[Euge]

Hi tmt,

First start with

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

Can you find this limit?

 

[Prove It]

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.

 

[Euge]

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 $\lim\limits_{r\to 0} r\cos(2\theta)$.

 

[Prove It]

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)

 

[tmt1]

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?

 

[Euge]

So it evaluates to 0?

Yes, that's correct.