Partial derivative using the definition

[U.Renko]

Homework Statement

So I'm supposed to find the partial derivatives and calculate them at point (0,2) using the definition

Homework Equations

$f(x,y) = x^2y\sin(1/x)$ IF x ≠ 0
$f(x,y) = 0$ IF x = 0

The Attempt at a Solution

$\frac{lim_{\Delta_x\rightarrow0} = (x + \Delta_x)^2y\sin(1/x+\Delta_x) - x^2y\sin(1/x)}{\Delta_x}$the thing is, I'm not very sure how to solve
$\lim_{\Delta_x\rightarrow0} \sin(1/x+\Delta_x)$

and if I just consider it the same as sin(1/x)
things get messed up

also this is my first time using LaTex.
Hope I did fine.

 

[mighty2000]

Hi there,

To solve the limit \lim_{\Delta_x\rightarrow0} \sin(1/x+\Delta_x), you can use the following property of limits:

\lim_{x\rightarrow a} (f(x) + g(x)) = \lim_{x\rightarrow a} f(x) + \lim_{x\rightarrow a} g(x)

Therefore, you can split the limit into two separate limits:

\lim_{\Delta_x\rightarrow0} \sin(1/x+\Delta_x) = \lim_{\Delta_x\rightarrow0} \sin(1/x) + \lim_{\Delta_x\rightarrow0} \sin(\Delta_x)

Since \lim_{\Delta_x\rightarrow0} \sin(\Delta_x) = 0, the second limit becomes 0. So now you just have to find the first limit, which you can do by using the squeeze theorem, as follows:

-1 \leq \sin(1/x) \leq 1

Therefore, by the squeeze theorem, \lim_{\Delta_x\rightarrow0} \sin(1/x) = 0.

So the final answer for the limit is 0.

I hope this helps! Let me know if you have any other questions.