1

What is the limit value of $\frac{sin(x)}{|x|}$ as $x\rightarrow 0?$

I tried to solve this question by dividing it into two cases $\mathrm{(i)} \ x>0,$ and $\mathrm{(ii)}\ x<0.$ When $x>0,$ the limit value goes to one, as we know.

But, when $x<0$ the absolute of x becomes $-x.$ So, the limit is -1.

So i conclude it that it doesn't have limit value because when $x>0$ it has 1 and when $x<0$ it has -1.

Am i correct, or there is a mistake?

user154797
  • 57
  • Yes, if the limit from the left does not match the limit from the right, the limit doesn't exist. – gammer Apr 02 '17 at 02:44
  • oh okay i am just not sure that they have different limit value –  Apr 02 '17 at 02:46
  • Why do you think it becomes -1 when x<0? –  Apr 02 '17 at 02:51
  • 1
    The absolute value will still positive even if x value is negative because you will take the absolute of the negative. –  Apr 02 '17 at 02:54
  • oh i think that when x<0 absolute x becomes -x so that negative values become positive value. and then we could take out the '-1' in front of limit f(x) and represent the function when x<0, '-lim sin(x)/x as x->0'. we knew that when lim sin(x)/x =1. so eventually the limit value becomes -1, for x<0 –  Apr 02 '17 at 03:22
  • @F.Alanazi, probably because sin is an odd function and so it determines the sign of $\sin(x)/|x|$, which is relevant when you're approaching zero from the left vs the right. – gammer Apr 02 '17 at 03:26
  • @gammer so your opinion is, my answer is right? –  Apr 02 '17 at 03:26
  • 1
    You never proved your assertion that the left hand limit was -1 and the right hand limit was 1. – gammer Apr 02 '17 at 03:28
  • yes, i didn't prove lim sin(x)/x =1. but i just want to check that my logic is correct... (x>0 we can just say lim sin(x)/x so the limit value is 1, and x<0 the function becomes lim sin(x)/-x so the limit value is -1...) –  Apr 02 '17 at 03:32
  • The |x| = x (always positive even if x is negative (-x). |-x|=|x|= x. So you can't say we take the - (negative) out the limit. –  Apr 02 '17 at 05:16
  • 2
    @F.Alanazi $|x|=-x $ when $x<0$. User 154797 is right. – Mark S. Apr 02 '17 at 07:59
  • @F.Alanazi bear in mind his is sin not cos ;-) – Hugh Perkins Apr 02 '17 at 12:51
  • @user154797 Please use mathjax. You can learn the basics here: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference –  Jan 16 '18 at 22:19

2 Answers2

1

From wikipedia:

"Alternatively x may approach p from above (right) or below (left), in which case the limits may be written as

$$ \lim _{x\to p^{+}}f(x)=L $$

or

$$\lim _{x\to p^{-}}f(x)=L$$

respectively. If these limits exist at p and are equal there, then this can be referred to as the limit of f(x) at p. If the one-sided limits exist at p, but are unequal, there is no limit at p (the limit at p does not exist). If either one-sided limit does not exist at p, the limit at p does not exist."

Therefore you are correct that the limit of $\sin(x) / \text{abs}(x)$ at $x=0$ does not exist.

Hugh Perkins
  • 687
  • 1
  • 7
  • 14
1

For those trying to understand how the author concluded how $$\lim_\limits{x\to0^-} \frac{\sin(x)}{|x|} = -1$$ and $$\lim_\limits{x\to0^+} \frac{\sin(x)}{|x|} = 1$$

and therefore $$\lim_\limits{x\to0} \frac{\sin(x)}{|x|} = \nexists$$

, I will assume that the reader recognizes $$\lim_\limits{x\to0} \frac{\sin(x)}{x} = 1 $$ Consider when we approach zero from the left side, the values are negative. When x is negative, $|x| = -x$ (e.g consider $x=1$). Thus,

$$\lim_\limits{x\to0^-} \frac{\sin(x)}{|x|} = \lim_\limits{x\to0^-} \frac{\sin(x)}{-x} = - \frac{\sin(x)}{x} = -1 $$ When we approach from the right side, $x>0$ and therefore positive. Thus, $$\lim_\limits{x\to0^+} \frac{\sin(x)}{x} = \lim_\limits{x\to0^+} \frac{\sin(x)}{x} = \frac{\sin(x)}{x} = 1 $$ Since they both exist but at different values, we must conclude that the limit does not exist ($\nexists$).

DFeng
  • 111