11
$\begingroup$

Why does this this limit not exist?

$$\lim_{x \to \infty} \arcsin \left({x+1\over x}\right)$$

According to me on dividing both the numerator and the denominator by $x$ and then putting $ x = \infty $ we should get $ \arcsin (1) $ which is equal to $ \frac{\pi} 2$ . Where am I wrong?

$\endgroup$
  • 2
    $\begingroup$ The expression $\lim_{x \to \infty} \arcsin \left({x+1\over x}\right)$ is meaningless since $\frac{x+1}{x} >1$. $\endgroup$ – user May 8 '18 at 17:06
  • 4
    $\begingroup$ Note that if $\arcsin$ is considered a function from $\Bbb C \to \Bbb C$, then the function is defined for arguments $> 1$, and the limit does exist. But whether it is $\pi/2$ depends on which branch of the $\arcsin$ function you choose. $\endgroup$ – Paul Sinclair May 8 '18 at 17:08
  • $\begingroup$ Before you take the limit, ask yourself: What is the domain of your function? Conclusion? $\endgroup$ – imranfat May 9 '18 at 0:24
11
$\begingroup$

The $\arcsin$ function is only defined on the domain $-1 \le x \le 1$. Since the input ${x+1 \over x} > 1 \,\forall x > 0$, the limit does not exist.

$\endgroup$
  • 2
    $\begingroup$ I wouldn't say that the limit doesn't exist, maybe it is more correct to say that the expression is not well-defined for x>0. $\endgroup$ – user May 8 '18 at 15:36
  • 4
    $\begingroup$ I prefer use the expression the limit doesn't exist when the expression is well defined but for example the function oscillates. $\endgroup$ – user May 8 '18 at 15:39
  • $\begingroup$ @gimusi I guess that wording is better since we're approaching the function from the right side which doesn't exist. $\endgroup$ – Andrew Li May 8 '18 at 16:05
  • $\begingroup$ It is of course a matter of convention and definition. I use the term "doesn't exist" for limit as for example $\lim_{x\to \infty} \sin x$ since the expression is well defined. In this case of that OP we can't even apply the definition of limit since the expression in not well-defined. $\endgroup$ – user May 8 '18 at 17:04
  • 3
    $\begingroup$ It is perfectly reasonable to say that the limit does not exist, since, well, no such limit exists. I think that reserving the phrase "does not exist" for only oscillatory functions is somewhat idiosyncratic. $\endgroup$ – Xander Henderson May 8 '18 at 22:39
23
$\begingroup$

Because $\frac{x+1}{x}>1$ for $x>0$, and $\arcsin{y}$ is not defined for $y>1$.

On the other hand, the limit as $x \to -\infty$ does exist, since $-1<\frac{x+1}{x}<1$ for sufficiently large negative $x$, and is $\pi/2$.

$\endgroup$
  • 3
    $\begingroup$ Most complete answer, IMHO. $\endgroup$ – user May 8 '18 at 15:40
4
$\begingroup$

The limit does not exist, because $\frac{x+1}{x}$ approaches $1$ from the right, where $\arcsin(x)$ is not defined.

$\endgroup$
0
$\begingroup$

...on dividing both the numerator and the denominator by x and then putting $x=\infty$...

if you could just do that, then there wouldn't really be a need for ever using limits. $\infty$ is not a number (in standard analysis, that is), so you can't “put $x=\infty$”.

Instead, the whole idea of the limit is to put in ever larger finite values for $x$ and still get a result that's not only always finite, but actually converges towards some point (which we then call the limit). This does work for $$ \lim_{x\to\infty} \arcsin\Bigl(\underbrace{\frac{x}{x+1}}_{y}\Bigr) $$ because here, you always have $0<y<1$, so can always find a solution to $y = \sin t$, and because $y$ goes asymptotically to $1$, this converges to a single point:

Convergent limit

But it doesn't at all work for the limit you're asking about, because here $y>1$ for a finite $x$, and that means you don't actually ever get a solution at all. Thus there also can't be a limit.

Non-existant limit


It only converges if you actually choose always the same solution, such as always the absolute-smallest one, which is what the $ⅹcsin$ function yields.

$\endgroup$
0
$\begingroup$

You are wrong to assume that

$$\lim_{x\to a} f(x)=f(\lim_{x\to a}x).$$

You have a perfect counter-example before you.

$\endgroup$
  • $\begingroup$ This downvote is undue. I do answer the question "Where am I wrong". $\endgroup$ – Yves Daoust May 10 '18 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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