Finding the value of an equation by using its two forms?

[Frigus]

In Resnick halliday book during finding capacitance of isolated sphere they used equation of spherical capacitor[4πε₀(ab)/b-a,where a is inner radius and b is outer radius.] And took b common and equation becames 4πε₀(a)(1-a/b) and then they put radius of outer sphere infinity and then a/b becomes 0 and equation comes out to be 4πε₀a but if we put b infinity in 1st equation [4πε₀(ab)/b-a] then we can't solve it as infinity over infinity is not 1 but undefined.
Thanks.

 

[fresh_42]

Your question comes down to: Why is $\lim_{n \to \infty}\dfrac{n}{n^2}=0$?

We could formally prove it.

Or just think of the $n$ all being natural numbers which can be cancelled. There is no need to divide infinities.

The third possibility is to plot the function $n \longmapsto \dfrac{n}{n^2}$ which will turn out to be the standard hyperbola $n \longmapsto \dfrac{1}{n}$.

Which of the above would convince you?

 

[Frigus]

Your question comes down to: Why is limn→∞nn2=0?

I can't understand how this relates to my question.
And it makes sense that $\frac 1 n$ approaches 0 as n approaches infinity

 

[fresh_42]

I can't understand how this relates to my question.
And it makes sense that $\frac 1 n$ approaches 0 as n approaches infinity

It is the same principle, I only left out all the factors.
\begin{align*}
\lim_{b \to \infty}\dfrac{4\pi\varepsilon_0(ab)}{b-a}&=\lim_{b \to \infty}4\pi\varepsilon_0a\cdot\dfrac{b}{b}\cdot\dfrac{1}{\dfrac{b}{b}-\dfrac{a}{b}}\\
&=4\pi\varepsilon_0a\cdot \lim_{b \to \infty}\dfrac{1}{1-\dfrac{a}{b}}\\
&=4\pi\varepsilon_0a\cdot \dfrac{1}{1-0}\\
&=4\pi\varepsilon_0a
\end{align*}
$\dfrac{1}{1-\dfrac{a}{b}}$ does not equal $1$, but its limit for $b\to \infty$ does.

 

[Frigus]

It is the same principle, I only left out all the factors.
\begin{align*}
\lim_{b \to \infty}\dfrac{4\pi\varepsilon_0(ab)}{b-a}&=\lim_{b \to \infty}4\pi\varepsilon_0a\cdot\dfrac{b}{b}\cdot\dfrac{1}{\dfrac{b}{b}-\dfrac{a}{b}}\\
&=4\pi\varepsilon_0a\cdot \lim_{b \to \infty}\dfrac{1}{1-\dfrac{a}{b}}\\
&=4\pi\varepsilon_0a\cdot \dfrac{1}{1-0}\\
&=4\pi\varepsilon_0a
\end{align*}
$\dfrac{1}{1-\dfrac{a}{b}}$ does not equal $1$, but its limit for $b\to \infty$ does.

But how can we put infinity value in equation 4πε₀$\frac {ab}{a-b}$ directly,
If we do so then it would become
4πε₀$\frac {∞a}{∞-a}$ and we know ∞ × some number is ∞ and ∞-some number is again infinity and we know $\frac ∞ ∞$ is undefined.
Can we solve this equation by just putting ∞ directly?
Thanks

 

[fresh_42]

Can we solve this equation by just putting ∞ directly?

No. This would lead to an expression which is simply not defined. What you can do is use epsilontics to prove the limit, but even then you will probably cancel $b$ for the sake of simplicity.

 

[Delta2]

We have an indeterminate form $\frac{\infty}{\infty}$ as you say but in this specific case we can remove the indeterminacy either by doing some neat algebraic tricks as @fresh_42 does at post #4, or by using the heavy machinery against indeterminacies like the L'Hospital rule, or even use epsilon-delta method. With either way we can remove the indeterminacy and find the limit which is 1 in this case.

This doesn't mean that all the indeterminacies of the form $\frac{\infty}{\infty}$ have limit 1, just this specific one. Other indeterminacies of the same form might have different limit or the limit might not exist, like for example $\frac{n+a}{n^2}$ which has limit 0, or $(-1)^n\frac{n^2}{n+a}$ in which the limit doesn't exist.

 

[Merlin3189]

May I as a naive non-mathematician make a suggestion to Hemant?

But how can we put infinity value in equation 4πε₀$\frac {ab}{a-b}$ directly,

We can't, because ∞ is not a number. It is more a symbol for an idea.

I think (as a non-mathematician) that when someone writes ∞ into an an expression as if it were a number, they are really using it as a shorthand for saying, "the limit of the expression as this number grows very large".
So you get into trouble if you read it as an actual number and start trying to do arithmetic with it.
You can only get away with the shorthand when a simple obvious limit does exist, like $\frac1n$ as n->∞.
If it gives you something dubious, like your $\frac{\infty}{\infty}$ then you need to do a bit of manipulation to get something(s) that you can find the limit(s) of.

For me, evaluating something "as n tends to infinity", means working out where that expression is going as n gets bigger. There are tests and rules to make sure it is really going somewhere definite and that you can get as close as you like to the destination.
But there is not a number "infinity" that makes the expression equal to the limit.

 

[Frigus]

May I as a naive non-mathematician make a suggestion to Hemant?

If you are a naive non-mathematician then I think I would be an toddler.

I think (as a non-mathematician) that when someone writes ∞ into an an expression as if it were a number, they are really using it as a shorthand for saying, "the limit of the expression as this number grows very large".
So you get into trouble if you read it as an actual number and start trying to do arithmetic with it.
You can only get away with the shorthand when a simple obvious limit does exist, like 1n as n->∞.
If it gives you something dubious, like your ∞∞ then you need to do a bit of manipulation to get something(s) that you can find the limit(s) of.

For me, evaluating something "as n tends to infinity", means working out where that expression is going as n gets bigger. There are tests and rules to make sure it is really going somewhere definite and that you can get as close as you like to the destination.
But there is not a number "infinity" that makes the expression equal to the limit.

Earlier I thought I have to remember it just as a bare fact but now after this post it makes a lot of sense.
Thanks.