Lim x->+inf xsin(1/x)

[fishingspree2]

Hello,

$$\[ \mathop {\lim }\limits_{x \to + \infty } x\sin ({\textstyle{1 \over x}}) \]$$

In my textbook they show a way to do this limit using variable substitution. I understand how they did it,

however, before I saw what they did, I tryed to work it out intuitively... as x goes to infinity, 1/x goes to 0... since sin(0) = 0, then the limit must be equal to 0

the correct answer is 1, but I don't understand why the intuitive method fails. can somebody help me out?

thank you, sorry for bad english

 

[HallsofIvy]

Hello,

$$\[ \mathop {\lim }\limits_{x \to + \infty } x\sin ({\textstyle{1 \over x}}) \]$$

In my textbook they show a way to do this limit using variable substitution. I understand how they did it,

however, before I saw what they did, I tryed to work it out intuitively... as x goes to infinity, 1/x goes to 0... since sin(0) = 0, then the limit must be equal to 0

"Intuitively" sin(1/x) goes to 0. It does not follow that x sin(1/x) goes to 0 because x is going to infinity. You can't say "$\infty*0= 0$".

the correct answer is 1, but I don't understand why the intuitive method fails. can somebody help me out?

thank you, sorry for bad english

Your English is excellent.

 

[ElectroPhysics]

Apply the La'Hospital rule

for that first of all convert the equation to form such that after applying limit directly we get 0/0 or infinity/infinity form. Then differentiate both the numerator and the denomenator and then apply the limit thus

f(x) = xsin(1/x) convert to f(x)/g(x) form i.e.
f(x)/g(x) = sin(1/x)/1/x which is now in the form of 0/0 ,then according to La'Hospital rule
f(x)/g(x) = f'(x)/g'(x) thus
f'(x)/g'(x) = [-1/x²cos(1/x)]/-1/x²

this gives

f'(x)/g'(x) = cos(1/x) now apply the limit to this derivative
which gives cos(1/infinity) = cos(0) = 1 which is the answer

 

[Ryker]

Sorry to bump such an old thread, but is there a way to show that the function x*sin(1/x) tends to 1 as x goes to infinity without using l'Hospital's rule?

 

[micromass]

Do you know the formula

$$\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1$$

if you don't know it, then I don't think you can do the question...

 

[Ryker]

I remember it from high school, but we haven't proven it in class at the university. We have done the limit of this function as x approaches 0, however, and with that one I don't have a problem. Is there perhaps a way you'd be able to squeeze xsin(1/x) between two functions that converge to 1 as x approaches plus infinity?

 

[micromass]

You could probably prove it with the following inequality

$$|\sin(1/x)|\leq 1/|x|\leq |\tan(1/x)|$$

The proof of this is nontrivial... You can see it in the following great video:

 

[avneell]

squuezing theorem

using squuezing theorem

limx→∞ tan-1(x)
x

solve.

 

[Ray Vickson]

Hello,

$$\[ \mathop {\lim }\limits_{x \to + \infty } x\sin ({\textstyle{1 \over x}}) \]$$

In my textbook they show a way to do this limit using variable substitution. I understand how they did it,

however, before I saw what they did, I tryed to work it out intuitively... as x goes to infinity, 1/x goes to 0... since sin(0) = 0, then the limit must be equal to 0

the correct answer is 1, but I don't understand why the intuitive method fails. can somebody help me out?

thank you, sorry for bad english

For small |y|, $\sin y = y + O(|y|^3),$ so for large x, $\sin(1/x) = 1/x + O(1/x^3).$ That implies $x \sin(1/x) = 1 + O(1/x^2) \rightarrow 1$ as $x \rightarrow \infty.$

RGV