Understanding the Limit of x * cot(x) as x Approaches 0: Explained and Solved

  • Thread starter Thread starter uman
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary
SUMMARY

The limit of x * cot(x) as x approaches 0 is 1, not 0, as demonstrated through the correct application of limit properties. The misconception arises from incorrectly assuming that the limit can be separated into the product of individual limits, which is not valid when one of the limits approaches infinity. The correct approach involves recognizing that cot(x) approaches infinity as x approaches 0, and using the identity x * cot(x) = (x/sin(x)) * cos(x), where the limit of x/sin(x) approaches 1 and cos(x) approaches 1. Thus, lim x->0 x * cot(x) = 1.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with trigonometric functions, specifically cotangent
  • Knowledge of L'Hospital's Rule
  • Basic concepts of continuity in functions
NEXT STEPS
  • Study the application of L'Hospital's Rule in limit problems
  • Learn about the behavior of trigonometric functions near their limits
  • Explore the concept of indeterminate forms in calculus
  • Review the continuity of functions and their limits
USEFUL FOR

Students of calculus, mathematics educators, and anyone looking to deepen their understanding of limits and trigonometric functions in calculus.

uman
Messages
348
Reaction score
1
Hi all. I encountered the following problem:

lim x->0 x * cot(x). My calculator gives the value of 1 for this limit, which can be corroborated by looking at the graph of y = x * cot(x).

However, I reason that it should be zero, because:

lim x->0 x * cot(x) = (lim x->0 x) * (lim x->0 cot(x))
= 0 * lim x->0 cot(x)
= 0 (because anything multiplied by zero is zero.)

Clearly, my reasoning is flawed. What is wrong with it? And what is the correct way to do this problem?

Thank you.
 
Physics news on Phys.org
uman said:
lim x->0 x * cot(x) = (lim x->0 x) * (lim x->0 cot(x))

In general, \lim_{x\to a} f(a)g(a) \neq (\lim_{x\to a} f(a)) (\lim_{x\to a} g(a))

For example, as x approaches zero, (x/x) approaches 1. If you split it like you did, we get zero times infinity. Clearly "anything multiplied by zero is zero." does not hold.
 
it's as Gib Z says, but just so you are clear, lim fg = limf limg IF limf and limg both exist, and the same goes for sums, lim(f + g) = limf + limg if both limf and limg exist, this is a common mistake. so think about this so you don't make it again
 
Last edited:
uman said:
Hi all. I encountered the following problem:

lim x->0 x * cot(x). My calculator gives the value of 1 for this limit, which can be corroborated by looking at the graph of y = x * cot(x).

However, I reason that it should be zero, because:

lim x->0 x * cot(x) = (lim x->0 x) * (lim x->0 cot(x))
= 0 * lim x->0 cot(x)
= 0 (because anything multiplied by zero is zero.)
That statement is wrong: 0 multiplied by \infty is not 0! It is correct that 0 multiplied by any real number is 0 but that limit is not a real number. The others said that "lim x->0 cot(x)" does not exist which is better wording for the same thing.

Clearly, my reasoning is flawed. What is wrong with it? And what is the correct way to do this problem?

Thank you.

cot(x)= cos(x)/sin(x). x cot(x)= x cos(x)/sin(x)= (x/sin(x))(cos(x)). It is proved in Calculus that \lim_{x\rightarrow 0} sin(x)/x= lim_{x\rightarrow 0}x/sin(x)= 1. Since cos(x) is continuous and cos(0)= 1, the limit of x cot(x)= 1.
 
You can use L'Hospital's rule for x/tan(x) goes to 1/(secx)^2 upon derivation top and bottom, which goes to 1 as x goes to 0.
 
Last edited:
Thanks a million, both of you!
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

Similar threads

Replies
6
Views
2K
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
993
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K