1. ## limit definition

Let $\displaystyle f(x,y)=sin|x|cos|y|$.

(a) Show from the limit definition that $\displaystyle f_y(0,0)$ exists, and find its value.

(b) Show from the limit definition that $\displaystyle f_y(0,0)$ does not exist.

My Attempt:

(a)
$\displaystyle \lim_{\Delta y \to 0} \frac{f(x_0, y_0 + \Delta y)-f(x_0,y_0)}{\Delta x}$

$\displaystyle \lim_{\Delta y \to 0} \frac{sin|0+|cos|0+\Delta x|-sin|0|xos|0|}{\Delta y}$

$\displaystyle \lim_{\Delta y \to 0} \frac{0}{\Delta y} = 0$

Is the value zero?

(b)
I can't prove that the limit doesn't exits because it turns out it exists:

$\displaystyle \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)-f(x_0,y_0)}{\Delta x}$

$\displaystyle \lim_{\Delta x \to 0} \frac{sin|0+ \Delta x|cos|0|-sin|0|cos|0|}{\Delta x}$

$\displaystyle \lim_{\Delta x \to 0} \frac{sin|\Delta x|}{\Delta x} = 1$

(since limit as x ->0 sinx/x=1)

I'm confused can anyone help?

2. Originally Posted by demode
Let $\displaystyle f(x,y)=sin|x|cos|y|$.

(a) Show from the limit definition that $\displaystyle f_y(0,0)$ exists, and find its value.

(b) Show from the limit definition that $\displaystyle f_y(0,0)$ does not exist.

My Attempt:

(a)
$\displaystyle \lim_{\Delta y \to 0} \frac{f(x_0, y_0 + \Delta y)-f(x_0,y_0)}{\Delta x}$

$\displaystyle \lim_{\Delta y \to 0} \frac{sin|0+|cos|0+\Delta x|-sin|0|xos|0|}{\Delta y}$

$\displaystyle \lim_{\Delta y \to 0} \frac{0}{\Delta y} = 0$

Is the value zero?
Yes, that's correct.

(b)
I can't prove that the limit doesn't exits because it turns out it exists:

$\displaystyle \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)-f(x_0,y_0)}{\Delta x}$

$\displaystyle \lim_{\Delta x \to 0} \frac{sin|0+ \Delta x|cos|0|-sin|0|cos|0|}{\Delta x}$

$\displaystyle \lim_{\Delta x \to 0} \frac{sin|\Delta x|}{\Delta x} = 1$

(since limit as x ->0 sinx/x=1)

I'm confused can anyone help?
But you do not have $\displaystyle \lim_{x\to 0}\frac{sin x}{x}$, you have $\displaystyle \lim_{x\to 0}\frac{sin|x|}{x}$

Look at the two one sided limits, as x goes to 0 from above and as x goes to 0 from below, remembering that for x< 0, sin(|x|)= sin(-x)= - sin(x).

3. But you do not have $\displaystyle \lim_{x\to 0}\frac{sin x}{x}$, you have $\displaystyle \lim_{x\to 0}\frac{sin|x|}{x}$

Look at the two one sided limits, as x goes to 0 from above and as x goes to 0 from below, remembering that for x< 0, sin(|x|)= sin(-x)= - sin(x).
Could you explain a little bit more please because I'm not sure if I really understand this. We are not concerned with x<0, we only care about (0,0), and as x goes to 0, sin|0|=0 and that makes the limit zero!