Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\lim_{x \to \infty}\frac{\sin 2x}{4x}$$

My book says:

Start by examinig the numerator of the given function, $\sin 2x$.
The $\sin$ function has a minimum absolute value of $0$ and a maximum absolute value of $1$.
Thus, the range of the absolute value of $\sin 2x$ is:

$$0 \leq |\sin 2x| \leq 1.$$

Divide each part of the inequality by $4x$:
$$0 \leq |\frac{\sin 2x}{4x}| \leq \frac{1}{4x}.$$


My question is:

1) Why do we use these absolute values? Why not squeeze $\sin 2x$ between $-1$ and $1$? Isn't $[-1,1]$ the range of the $\sin$ function? Why are we considering the range of the absolute value?

2) Can we squeeze the whole function $\frac{\sin 2x}{4x}$ between two values, why does it concentrate on just the numerator, $\sin 2x$?

Thank you.

share|improve this question
    
I changed multiple instances of \textrm{sin} to \sin, which is standard usage. If you write a\textrm{sin}b, you don't get proper spacing, whereas with a\sin b you do: $a\textrm{sin}b$ versus $a\sin b$. –  Michael Hardy Nov 12 '13 at 2:55
    
@MichaelHardy I see, thanks. –  user437158 Nov 12 '13 at 2:57

1 Answer 1

up vote 0 down vote accepted

Succint answers to your questions:

1)

  1. When having limits that oscilate between negative and positive values, it is a common technique to use absolute values.
  2. Yes you can do so!
  3. Yes it is.
  4. This is the same question as the first one

2)

  1. Because $\sin2x$ is problematic to deal with, but $\frac{1}{x}$ is not.
share|improve this answer
    
Thanks. Can you expand a little on why $\sin 2x$ is problematic, but $\frac{1}{x}$ is not? Is it because it oscillates? –  user437158 Nov 12 '13 at 3:10
    
@user436158 And also because it is non-constant, so it gives more freedom to the limit. When we bound it with constants, we get a much simpler expression that we can handle easily. –  chubakueno Nov 12 '13 at 3:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.