# Non-inverting op-amp amplifier gain as open-loop gain goes to infinity (limit calculation)

Why is the following true?

I can't seem to understand why. Shouldn't it just be infinity/1+infinity which is approximately 1?

It's comparing the vo/vin of a non-inverting amplifier with finite gain with that of one with infinite gain.

• infinity/(1+infinity) is an indeterminate form, it isn't equal to 1. If you get an indeterminate form it means you haven't found a solution and you need to try some other approach. – user253751 Mar 25 '18 at 23:06

\begin{align*} &= \lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}}\\\\ &= \lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}}\cdot\frac{\frac{1}{A}}{\frac{1}{A}}\\\\ &= \lim_{A\rightarrow\infty}\frac{1}{\frac{1}{A}+\frac{R_1}{R_1+R_F}}\\\\ &= \frac{1}{\frac{R_1}{R_1+R_F}}\\\\ &= \frac{R_1+R_F}{R_1}\\\\ &= 1+\frac{R_F}{R_1} \end{align*}

• Perfect. Thank you. – AlfroJang80 Mar 24 '18 at 21:07

This is of infinity/infinity form. So we will do a work around to calculate this limit. $$\frac{A}{1+\frac {AR_i}{R_i+R_f}} = \frac{A}{A(\frac{1}{A}+\frac {R_i}{R_i+R_f})} = \frac{1}{\frac{1}{A}+\frac {R_i}{R_i+R_f}}$$

Now you can apply limits.

$$\lim_{A\to \infty} \frac{1}{\frac{1}{A}+\frac {R_i}{R_i+R_f}} = \frac{1}{0+\frac {R_i}{R_i+R_f}} =\frac {R_i+R_f}{R_i}$$

• Perfect. Thank you. I think this is the same approach as jonk's. – AlfroJang80 Mar 24 '18 at 21:07

We can write your equation in a slightly different way $$A_{CL} = \frac{A}{1 + Aβ}$$

Where $β = \frac{R_1}{R_1 + R_F}$

And now if we divide this by $A$ we are going to get this:

$$A_{CL} = \frac{1}{(1/A) + β}$$

So, now $A$ is approaching the infinity $(1/A = 0)$

we can see that the closed loop gain is equal to:

$\Large \frac{1}{\beta} =\frac{R_1+ R_F}{R_1} =\frac{R_1}{R_1}+\frac{R_F}{R_1}= 1 +\frac{R_F}{R_1}$

A less rigorous method is to look at the denominator, and notice that, as A gets very large, A(R1/(R1 + RF)) gets much larger than 1, so the 1 can be discarded.

Then the ratio is easily evaluated and the A's drop out.

$\infty/\infty$ is an indeterminate form, so use L'Hôpital's rule

$$\lim_{A\rightarrow\infty}\frac{A}{1+A\frac{R_1}{R_1+R_F}} = \frac{\frac{d}{dA}(A)}{\frac{d}{dA}(1+A\frac{R_1}{R_1+R_F})} = \frac{1}{\frac{R_1}{R_1+R_F}} = 1 + \frac{R_F}{R_1}$$