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

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.

  • $\begingroup$ 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 $a\sin b$. $\endgroup$ Nov 12, 2013 at 2:55
  • $\begingroup$ @MichaelHardy I see, thanks. $\endgroup$
    – Emi Matro
    Nov 12, 2013 at 2:57

1 Answer 1


Succint answers to your questions:


  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


  1. Because $\sin2x$ is problematic to deal with, but $\frac{1}{x}$ is not.
  • $\begingroup$ Thanks. Can you expand a little on why $\sin 2x$ is problematic, but $\frac{1}{x}$ is not? Is it because it oscillates? $\endgroup$
    – Emi Matro
    Nov 12, 2013 at 3:10
  • $\begingroup$ @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. $\endgroup$
    – chubakueno
    Nov 12, 2013 at 3:24

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