# Why tan x=x as x approaches 0?

• I
Hi! In one of my textbook i saw the relation tan(x) = x where x is very small value and expressed in radians. I want to know why its true and how it actually works. I would appreciate someone's help

mfb
Mentor
Did you draw a sketch? A small part of the circle can be approximated by a straight line.
It is similar to the approximations ##\cos(x) \approx 1## and ##\sin(x) \approx x## for small x, and you can derive the approximation of the tangent that way as well.

fresh_42
Mentor
Hi! In one of my textbook i saw the relation tan(x) = x where x is very small value and expressed in radians. I want to know why its true and how it actually works. I would appreciate someone's help
One way to look at it is to take it as the linear approximation by the first derivative. We have ##\left. \dfrac{d}{dx}\right|_{x=0}\tan(x)=(1+\tan^2(x))|_{x=0}=1## which means that the tangent function is locally approximated by ##x \longmapsto (\tan(x))_0' \cdot x =1 \cdot x##.

Another way is to use the Taylor series at ##x=0## which is ##\tan(x) = x + O(x^3)\,.##

QuantumQuest
Did you draw a sketch? A small part of the circle can be approximated by a straight line.
It is similar to the approximations ##\cos(x) \approx 1## and ##\sin(x) \approx x## for small x, and you can derive the approximation of the tangent that way as well.
But why is it true for radians only? I solved on my calculator and this is what i saw :
Tan (0.12) = 0.12 (where 0.12 is radians)
But when x is in degrees
Tan(0.12) = 0.00209

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fresh_42
Mentor
But why is it true for radians only? I solved on my calculator and this is what i saw :
Tan (0.12) = 0.12 (where 0.12 is radians)
But when x is in degrees
Tan(0.12) = 0.00209
##\tan 0.12 \approx \tan 7° \approx [x + O(x^3)]_{at \, 0} = 0.12 \pm 0.002## which is close to ##0.12##.
##\tan 0.12° \approx \tan 0° \approx [x + O(x^3)]_{at \, 0} = 0 \pm 0## which is close to ##0.00209##.

You simply compare a value ##0.12## with its ##60-##fold value. But both are still in a very good approximation to ##\tan (x) \approx x##. Radians are the natural unit here, degree more because of historical reasons, habit and clarity for humans. The approximation ##\tan (x) \approx x## requires radians if taken numerically disregarding the units.

rishi kesh
TeethWhitener
Gold Member
But why is it true for radians only?
The series expansion of ##\tan x## is ##\tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5+\cdots##, where the powers of ##x## get bigger and bigger. This means for any value of ##x > 1##, the higher power terms in the series will contribute proportionally more than the lower terms. So the approximation only really works well when ##x\ll 1##, because then the higher order terms die out quickly. If you use degrees instead of radians, then you're effectively using ##d = \frac{180}{\pi}x## and calculating ##\tan d## instead of ##\tan x##. Since ##\frac{180}{\pi}\approx 57.2##, you should expect to get ##\tan d \approx \frac{1}{57.2}d##. Doing a quick calculation:
$$0.12°\times \frac{1}{57.2} \approx 0.002098$$
in line with what you would expect.

Mark44
Mentor
Hi! In one of my textbook i saw the relation tan(x) = x where x is very small value and expressed in radians.
No reputable book would say that tan(x) = x. The proper relationship is that ##\tan(x) \approx x## for values of x near 0 (and in radians).

Khashishi
But why is it true for radians only?
Radians are the natural units for angles.

If you take a circle with radius 1, then the circumference is ##2\pi##. And there are also ##2\pi## radians in a revolution. So, the arc length of a circle radius 1 is equal to the angle in radians. The tangent function is defined as the ratio of lengths of a right triangle ##\frac{opposite}{adjacent}##. If you take a right triangle with a small angle, the length of the opposite leg is very close to the arc length of the circle next to it, and the length of the adjacent leg is very close to the radius of the circle. So, the tangent is very close to the angle in radians.

Chestermiller, FactChecker and olivermsun
One can show using the L'Hopital's rule that
$$\lim_{x\to 0} \frac{\tan x}{x} = \lim _{x\to 0} \frac{1}{\cos ^2x} = 1$$
This immediately implies ##\tan x = x + \alpha (x) ##, where ##\alpha (x)\to 0 ## as ##x\to 0 ##. In other words, the closer you get to ##0 ##, the smaller the difference between ##\tan x ## and ##x ## becomes.

mfb
Mentor
This immediately implies ##\tan x = x + \alpha (x) ##, where ##\alpha (x)\to 0 ## as ##x\to 0 ##. In other words, the closer you get to ##0 ##, the smaller the difference between ##\tan x ## and ##x ## becomes.
That is a much weaker statement than the ratio. It would also apply for ##2x=x+\alpha(x)##, for example, but approximating 2x as x is usually a bad idea.

It implies ##\tan x = x(1 + \alpha (x)) ##, where ##\alpha (x)\to 0 ## as ##x\to 0 ##

QuantumQuest and nuuskur
Our statements are equivalent, although I like yours more as it is more explicit in a way.

mfb
Mentor
The statements are not equivalent. Your second statement just says tan(x) and x have the same limit for x->0, that is a much weaker statement.

The ratio is a strong statement, my reply was commenting on the remark afterwards only.

fresh_42
If ##\tan x = x + \alpha (x)##, then ##\frac{\tan x}{x} = 1 + o(x) ##. Conversely, if ##\lim_{x\to 0} \frac{\tan x}{x} = 1 ##, then ##\tan x = x(1+\alpha (x)) =: x + \hat{\alpha} (x) ##. We just label things differently, it seems.

Besides, if their limits are the same in the viewed process, the ratio statement follows (in this case).

mfb
Mentor
If ##\tan x = x + \alpha (x)##, then ##\frac{\tan x}{x} = 1 + o(x) ##.
That argument does not work in general, see my example with 2x instead of tan(x).
Conversely, if ##\lim_{x\to 0} \frac{\tan x}{x} = 1 ##, then ##\tan x = x(1+\alpha (x)) =: x + \hat{\alpha} (x) ##. We just label things differently, it seems.
That direction is fine but that is a much weaker statement on the right side if you just require ##\hat \alpha(x)## to go to 0.

What you need to make the two statements equivalent is the condition that ##\displaystyle \frac{\alpha(x)}{x} \to 0## instead of ##\alpha(x) \to 0##.

nuuskur
You are correct. I should have made my remainder term more explicit, hence my remark why I like yours more.

fresh_42
Mentor
You are correct. I should have made my remainder term more explicit, hence my remark why I like yours more.
Your mistake goes deeper than sloppiness, because your concept of a (linear) approximation by the first derivative missed the point. It is essential that the normed direction tends towards zero not just the remainder term. We have ##\tan(0+v) = \tan(0) + \tan'(0)\cdot v + r(v)## and ##\lim_{v \to 0}\dfrac{r(v)}{||v||} =0##. This is the reason why @mfb's counterexample works, if this is not the case. You cannot skip the denominator.

nuuskur
Radians are the natural units for angles.

If you take a circle with radius 1, then the circumference is ##2\pi##. And there are also ##2\pi## radians in a revolution. So, the arc length of a circle radius 1 is equal to the angle in radians. The tangent function is defined as the ratio of lengths of a right triangle ##\frac{opposite}{adjacent}##. If you take a right triangle with a small angle, the length of the opposite leg is very close to the arc length of the circle next to it, and the length of the adjacent leg is very close to the radius of the circle. So, the tangent is very close to the angle in radians.
This is the explanation i actually needed. But i want you to extend it a little bit, hopefully you will clear my doubt(please check my attachment). I will appreciate further reply from you :)

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Mark44
Mentor
This is the explanation i actually needed. But i want you to extend it a little bit, hopefully you will clear my doubt(please check my attachment). I will appreciate further reply from you :)
The photo you posted is the reason that we discourage images of work.
1. The image is unreadable because it is so small.
2. The image is rotated, making it difficult to read even if it were larger.

Khashishi