0
$\begingroup$

Like this equation :

$$\lim _{x\to 0 }x\sin \frac{1}{x}$$

$x$ is the denominator, but we can nevertheless plug $x=0$ into the function to get the limit, why?

$\endgroup$
3
  • 4
    $\begingroup$ This is not true, we can't "plug $x-0$ into the function" because the function is not defined at $x=0$. To show the limit is $0$ we have to use theorems on limits. $\endgroup$ Commented Sep 17, 2021 at 15:02
  • 1
    $\begingroup$ "but we can nevertheless plug x=0 into the function to get the limit" Says who? We can NOT do that. $\endgroup$
    – fleablood
    Commented Sep 17, 2021 at 17:02
  • $\begingroup$ We can use the squeeze theorem though. $-1 \le \sin \frac 1x \le 1$ so if $x > 0$ then $-x= -|x| \le x\sin\frac 1x \le x=|x|$ and if $x < 0$ then $-1\times x=-x = |x| \ge x\sin\frac 1x \ge 1\times x = x = -|x|$ so as long as $x \ne 0$ we have $-|x| \le x\sin \frac 1x \le |x|$. So $\lim_{x\to 0} -|x| \le \lim_{x\to 0} x\sin \frac 1x \le \lim_{x\to 0} |x|$ so $0 \le \lim_{x\to 0} x\sin \frac 1x \le 0$ so $\lim_{x\to 0} x\sin \frac 1x$. But we certainly can NOT plug $x=0$ into $x\sin \frac 1x$ to get $0 \times \frac 10 = ???????$ $\endgroup$
    – fleablood
    Commented Sep 17, 2021 at 17:12

2 Answers 2

2
$\begingroup$

As you state in the question itself, we find the limit of the function as $x$ APPROACHES $0$ - and not AT $0$. This means that we only need the function to be defined in the "immediate left" and the "immediate right" of $x$ = $0$.

Since we know with certainty that $sin(any real input)$ lies between $-1$ and $1$, when multiplied by a very, very small number (i.e. $x$ as it approaches $0$), the expression will also tend to $0$.

at 0

Recommended reading: https://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-1.html and the 3Blue1Brown video on Limits.

$\endgroup$
1
  • 1
    $\begingroup$ That's a post of my blog! Glad to know that you liked it. $\endgroup$
    – Paramanand Singh
    Commented Sep 17, 2021 at 20:05
0
$\begingroup$

You are never plugging in $x=0$ into the function. You are only looking at what happens when the value of $x$ approaches $0$, i.e., what happens to the value of the function as $x$ goes on decreasing. Note that $0$ may not be in the domain of the function, but every small real number (no matter how small it is) is in the domain, and these are the values that we are using to find the limit.

$\endgroup$
3
  • $\begingroup$ every small real number, not every small positive integer $\endgroup$
    – David P
    Commented Sep 17, 2021 at 15:57
  • $\begingroup$ @DavidP oh yes, true. That was what I meant. I don't know why I sometimes daydream and write those stupid things :| $\endgroup$ Commented Sep 17, 2021 at 15:58
  • $\begingroup$ @DavidP please check the edit. Is it fine now? $\endgroup$ Commented Sep 17, 2021 at 15:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .