Why does the lim Δx →0 change the ≈ to =?

[aclark609]

I was sifting through the beginning of my book when i came upon a section based on marginals and differentials. My question is why does Δy/Δx ≈ f'(x) when the lim Δx →0 Δy/Δx = f'(x)?


Δx = (x + Δx) - x ; therefore, Δy = f(x + Δx) - f(x) .

Δy/Δx = {f(x + Δx) - f(x)}/ Δx ≈ f'(x)

f'(x) = lim {f(x + Δx) - f(x)}/ Δx
Δx→0

In simplest terms, why does the lim Δx →0 change the ≈ to =?

 

[mfb]

f'(x) = lim {f(x + Δx) - f(x)}/ Δx
Δx→0

That is the definition of f'(x). It is true because it was defined that way.

 

[ahaanomegas]

I believe aclark609's specific question was about why the $\approx$ changes to a $=$ and the answer is that that's how limits are. Whilst calculating a limit, the approximate answer is made exact via an infinite number of precise approximations. That is exactly the beautiful spirit of Calculus. For example, $0.\overline{9} = 1$, it is not approximately $1$. If the number of $9$s after the $0$ was huge but finite, it would be approximately, and very close to, $1$. But, as the number of 0s grows more and more, the value becomes closer and closer to 1 - i.e. "approaching 1". The limit is the infinite case, which is exactly $1$.

Hope that helped!

 

[aclark609]

I believe aclark609's specific question was about why the $\approx$ changes to a $=$ and the answer is that that's how limits are. Whilst calculating a limit, the approximate answer is made exact via an infinite number of precise approximations. That is exactly the beautiful spirit of Calculus. For example, $0.\overline{9} = 1$, it is not approximately $1$. If the number of $9$s after the $0$ was huge but finite, it would be approximately, and very close to, $1$. But, as the number of 0s grows more and more, the value becomes closer and closer to 1 - i.e. "approaching 1". The limit is the infinite case, which is exactly $1$.

Hope that helped!

So by adding in the lim Δx →0 it pretty much replaces the ≈ with = because that is the whole concept of a limit, correct? If that is the case then wouldn't that make all differentials 'approximate.'

 

[Fredrik]

So by adding in the lim Δx →0 it pretty much replaces the ≈ with = because that is the whole concept of a limit, correct?

I wouldn't describe it that way. But the definition of the notation f'(x) does involve a limit.

If that is the case then wouldn't that make all differentials 'approximate.'

Depends on how you define the "differentials". You can e.g. define df as a function that takes two real numbers to a real number, like this: df(x,h)=f'(x)h for all x and all h. Now we have f'(x)=df(x,h)/h by definition. However, if you're asking whether
$$\frac{f(x+h)-f(x)}{h}$$ is equal to or approximately equal to f'(x), then the answer is that it's approximately equal to f'(x) when h is small.

 

[EuroNerd77]

It is the derivative of the curved line. y/x = slope. So Δy/Δx is just the "Calculus Way" of defining it.

 

[Office_Shredder]

So by adding in the lim Δx →0 it pretty much replaces the ≈ with = because that is the whole concept of a limit, correct? If that is the case then wouldn't that make all differentials 'approximate.'

I think your intuition is backwards here. It isn't that f'(x) is approximately Δy/Δx, and if you take the limit as Δx goes to zero you get an equality. It's that as Δx goes to zero, you get exactly the value f'(x) because that's how it's defined. The interesting point is that this implies Δy/Δx is going to approximate the value of f'(x) if Δx is small

 

[Fredrik]

It is the derivative of the curved line. y/x = slope. So Δy/Δx is just the "Calculus Way" of defining it.

x/y is the slope only when we're dealing with a function whose graph is a straight line through the origin.