Limit of the expression - with partial differentiated terms

D

Dinoduck94

New member
Joined
May 15, 2019
Messages
22
I have a coursework question that I'm hoping that I can get clarification on - please I don't want the answer, I just want clarification:

The question is below:
tma.png

The partial derivatives here confused me.
Am I supposed to treat this essentially as sin(delta_x)/delta_x and then find a trend as I make delta_x closer and closer to 0; or is there something that I'm supposed to do with the partial derivatives?

Thanks
 
Dinoduck94 said:
I have a coursework question that I'm hoping that I can get clarification on - please I don't want the answer, I just want clarification:

The question is below:
View attachment 14249

The partial derivatives here confused me.
Am I supposed to treat this essentially as sin(delta_x)/delta_x and then find a trend as I make delta_x closer and closer to 0; or is there something that I'm supposed to do with the partial derivatives?

Thanks
Click to expand...
I think that is the correct interpretation.
 
If this will make things easier for you, you can simply replace delta_x with x (or y or....)

I am also a bit confused here. If you know about partial derivative then you have taken calculus 1. Is this true?
If so, then you should have already seen this limit early in calculus 1 and with a good instructor you should have seen the geometric proof that limit equals ... well you asked not to be given the limit
 
Jomo said:
If this will make things easier for you, you can simply replace delta_x with x (or y or....)

I am also a bit confused here. If you know about partial derivative then you have taken calculus 1. Is this true?
If so, then you should have already seen this limit early in calculus 1 and with a good instructor you should have seen the geometric proof that limit equals ... well you asked not to be given the limit
Click to expand...

Hi Jomo,

I have taken Calculus 1, and have seen this expression when 'delta_x' was simply just 'x'.
I went further than what my material told me too and watched a video explaining the geometric proof, so as a result of both of these, I do know the result of the expression; however it was the partial derivatives in the expression that confused me!

Thanks
 
There are no "partial derivatives" there! The \(\displaystyle \delta x\) is not a "partial derivative". It is a way of saying "a small change in x".
 

Attachments

  • limit _{x-_0}( sin (x))_(x) - Wolfram_Alpha.pdf
    275.2 KB · Views: 5
Dinoduck94 said:
Hi Jomo,

I have taken Calculus 1, and have seen this expression when 'delta_x' was simply just 'x'.
I went further than what my material told me too and watched a video explaining the geometric proof, so as a result of both of these, I do know the result of the expression; however it was the partial derivatives in the expression that confused me!

Thanks
Click to expand...
Even if there was a partial derivative there (that is some function), since the function was going to 0 we have sin(p.d)/(p.d) approaching 1. (p.d is just some partial derivative)

For example, let f(x) = x^2 -x (which is a partial derivative of some function!). Then as f(x) approaches 0 we DO have sin(x^2-x))/(x^2-x) approaching 1
 
You must log in or register to reply here.
Top