How exactly is the value of $\frac{1}{y\sqrt{y^2+\frac{l^2}{4}}}$ as $y \gg l$ calculated?

Question

I am having trouble understanding if a particular calculation is or is not a limit calculation. I suspect it is not, but a particular set of notes from an MIT OCW physics course (problem starts on page 18, the limit is on page 20) seem to imply the calculation in question is a limit. The calculation is relatively simple, but I would really like to understand exactly what is being done to reach the result.

Note that I've slightly adapted the expressions to remove the parts that are specific to the physics domain (electromagnetism).

Consider the following expression for $E_p$:

$$E_p=\frac{1}{y\sqrt{y^2+\frac{l^2}{4}}}\tag{1}$$

I'd like to know what happens to this expression when $y$ is large relative to $l$.

The notes I am following state that "in the limit where $y \gg l$, the above expression reduces to the (point-charge) limit:"

$$E_p=\frac{1}{y^2}$$

As far as I can tell, if we take the limit of $(1)$ when $y \to \infty$ we get $0$.

I believe what is happening is that a linear approximation is being used in the denominator.

Here's what I came up with:

Rewrite $E_p$ $$E_p = \frac{1}{y\sqrt{y^2+\frac{l^2}{4}}}$$

$$=\frac{1}{y\sqrt{y^2(1+\frac{l^2}{4y^2})}}$$

$$=\frac{1}{y^2\sqrt{1+\frac{l^2}{4y^2}}}$$

Consider the term in the denominator $1+\frac{l^2}{4y^2}$

$$s=\frac{l}{2y}$$

$$1+(\frac{l}{2y})^2=1+s^2 = f(s)$$

$$y>>l \implies s \approx 0$$

I believe I can use a first order Taylor's expansion

$$\implies f(s) \approx f(0) +f'(0)s,\text{ near s=0}$$

$$\implies f(s) \approx 1$$

$$E_p \approx \frac{1}{y^2}$$

On the other hand I could have simply started by considering the limit of $E_p$ when $\frac{l}{y} \to 0$. In this case my question is: how does one calculate this limit? It's clear that the term in the square root goes to zero, but what should we say happens to the $y^2$ term?

Is this latter limit somehow connected or equivalent to the linear approximation I used, and was my calculation correct?

Answer 1

By Taylor,

$$\sqrt{y^2+1}=y\sqrt{1+\frac1{y^2}}=y\left(1+\frac1{2y^2}+o(y^{-2})\right)=y+\frac1{2y}+o(y^{-1})=y+o(1).$$

The correction term is negligible for large $y$.

Answer 2

Assuming $y>0$, you can rewrite your expression as $$ f(y)=\frac{1}{y^2}\Bigl(1+\frac{l^2}{4y^2}\Bigr)^{-1/2} $$ and we can use the Taylor expansion: $$ f(y)=\frac{1}{y^2}\Bigl(1-\frac{1}{2}\frac{l^2}{4y^2}+o(l^2/4y^2)\Bigr) $$ If $y\gg l$, you may decide to neglect also the second term inside the parentheses to find $f(y)\approx 1/y^2$; otherwise you may say that $$ f(y)\approx\frac{1}{y^2}-\frac{l^2}{8y^4} $$ or use more terms if appropriate.

Answer 3

(You asked how to take the limit as $l/y \rightarrow 0$.)

The "problem" with taking the limit as $l/y \rightarrow 0$ is that either $l$ is getting smaller, $y$ is getting larger, or both. Since we want to have an expression with $y$s and not $l$s, we need to arrange for all appearances of "$l$" to be in the combination "$l/y$". This is not initially the case.

Also, recall that $\sqrt{x^2} = |x|$ for any real $x$ (since the square root can only give you nonnegative values). \begin{align*} E_p &= \frac{1}{y\sqrt{y^2 + \frac{l^2}{4}}} \\ &= \frac{1}{y\sqrt{y^2 \left( 1 + \frac{l^2}{4y^2} \right)}} \\ &= \frac{1}{y|y|\sqrt{1 + \frac{l^2}{4y^2}}} \end{align*} Now, we have all instances of $l$ in $l/y$ combinations, so we can take the limit, leaving an expression in $y$s only. $$ \lim_{l/y \rightarrow 0} \frac{1}{y|y|\sqrt{1 + \frac{l^2}{4y^2}}} = \frac{1}{y|y|\sqrt{1 + \frac{0^2}{4}}} = \frac{1}{y|y|} \text{.} $$

And, actually, we want $y|y|$ here -- the $E$-field should be pointing away from the origin above and below the $x$-axis, not always pointing upwards. I don't see that we are working only on $y \geq 0$ and getting the lower half by reflection, so the usual explanation for why we get to confine our attention to $y \geq 0$ is missing.