Why does the limit of $x \tan(n/x)$ as $x$ approaches infinity always equal $n$?

Question

I was looking through this paper and found forward and backward transform equations for calculating radial distortion caused by fisheye lense abberation using the equations $f = \text{focal length}, r_u = \text{undistorted radius}, r_d = \text{distorted radius}$:

$r_d = f \arctan(\frac {r_u}{f})$

$r_u = f \tan(\frac {r_d}{f})$

I was previously confused by non polynomial representations of the distortion through there lack of explanation of how to plugin variables to the equation and get out a non distorted image (as if no abberation occured) like this one: in their words not mine...

$s =$ "a scalar"

$\lambda = $"controls the amount of distortion across the image"

$r_d = s \ln(1 + \lambda(r_u))$

$r_u = (e^\frac{r_d}s - 1)/\lambda$

I couldn't figure out what constants of lambda and s would get no distortion (this was important for programmic implementation and testing). Originally I didn't see how this was possible with the previous equation until I input higher and higher values of $f$, and I found out that the limit of $\tan(n/x)$ as $x$ approaches infinity always equal n, so you when $f = \infty, r_u = r_d$. (note the later equation set did not seem to have any consistency with inputs, so what ever makes $r_u = r_d$ isn't $0,1,\infty,$ or negation of any of those values)

I tried looking this up, and I was able to find this, however the logic used here cannot be applied for any numerator in tangent except 1, since $\sec^2(1/\infty) = \sec^2(0) = 1$ implies that $\sec^2(2/\infty)$ also $= 1$.

What are the steps to prove this/how would I even start formulating how this is true in my head?

EDIT: meant $x \tan(n/x)$, not $\tan(n/x)$, though you really shouldn't just be looking at the title and calling it a day :)

Answer 1

$$\lim_{x\to\infty}\tan\frac nx=0$$ because $$\lim_{x\to\infty}\frac nx=0.$$

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The true question is

$$\lim_{x\to\infty}x\tan\frac nx=n$$ because $$\lim_{x\to\infty}x\frac nx=n,$$ using the Taylor development to the first order.

Answer 2

Straight limit with L'Hopital: $$\lim_{x\to\infty}x\tan\Bigl(\frac{n}{x}\Bigr)=\lim_{x\to\infty}\dfrac{\tan\bigl(\frac{n}{x}\bigr)}{\frac{1}{x}}\to\frac{0}{0}\xrightarrow{\text{L'Hopital}}\lim_{x\to\infty}\dfrac{\tan\bigl(\frac{n}{x}\bigr)}{\frac{1}{x}}=\lim_{x\to\infty}\dfrac{n\sec^2\bigl(\frac{n}{x}\bigr)\cdot\left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}=\lim_{x\to\infty}\dfrac{n}{\cos^2\bigl(\frac{n}{x}\bigr)}=n$$ As others have said, $\lim\limits_{x\to\infty}\tan\Bigl(\frac{n}{x}\Bigr)=0$

Answer 3

The derivative of $\tan(z)$ at $z=0$ is $1$. That is what your fact says: $$ 1 = \tan'(0) = \lim_{h \to 0}\frac{\tan(h)-\tan(0)}{h} $$ so for fixed $n$, let $h=n/x$, so that $h \to 0$ as $x \to \infty$. Thus $$ 1 = \lim_{x \to \infty} \frac{\tan(n/x) - 0}{n/x} = \lim_{x \to \infty} \frac{x}{n}\tan\frac{n}{x} = \frac{1}{n} \lim_{x \to \infty} x\tan\frac{n}{x} $$ equivalent to your result.

Answer 4

Hint Set $h=\frac{n}{x}\to 0 $ as $x\to \infty.$ and use. $$\lim_{h\to 0}\frac{\tan h}{h} =1$$