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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?

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    $\begingroup$ The expression $\lim_{x \to \infty} \arcsin \left({x+1\over x}\right)$ is meaningless since $\frac{x+1}{x} >1$. $\endgroup$
    – user
    Commented May 8, 2018 at 17:06
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    $\begingroup$ 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. $\endgroup$ Commented May 8, 2018 at 17:08
  • $\begingroup$ Before you take the limit, ask yourself: What is the domain of your function? Conclusion? $\endgroup$
    – imranfat
    Commented May 9, 2018 at 0:24

5 Answers 5

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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 $\pi/2$.

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    $\begingroup$ Most complete answer, IMHO. $\endgroup$
    – user
    Commented May 8, 2018 at 15:40
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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.

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    $\begingroup$ 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
    Commented May 8, 2018 at 15:36
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    $\begingroup$ I prefer use the expression the limit doesn't exist when the expression is well defined but for example the function oscillates. $\endgroup$
    – user
    Commented May 8, 2018 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
    Commented May 8, 2018 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
    Commented May 8, 2018 at 17:04
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    $\begingroup$ 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
    Commented May 8, 2018 at 22:39
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The limit does not exist, because $\frac{x+1}{x}$ approaches $1$ from the right, where $\arcsin(x)$ is not defined.

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...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.

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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.

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  • $\begingroup$ This downvote is undue. I do answer the question "Where am I wrong". $\endgroup$
    – user65203
    Commented May 10, 2018 at 13:07

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