# Why does the limit of $\lim_{x \to \infty} \arcsin \left(\frac{x+1}{x}\right)$ not exist?

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?

• The expression $\lim_{x \to \infty} \arcsin \left({x+1\over x}\right)$ is meaningless since $\frac{x+1}{x} >1$. – user May 8 '18 at 17:06
• 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. – Paul Sinclair May 8 '18 at 17:08
• Before you take the limit, ask yourself: What is the domain of your function? Conclusion? – imranfat May 9 '18 at 0:24

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.

• 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. – user May 8 '18 at 15:36
• I prefer use the expression the limit doesn't exist when the expression is well defined but for example the function oscillates. – user May 8 '18 at 15:39
• @gimusi I guess that wording is better since we're approaching the function from the right side which doesn't exist. – Andrew Li May 8 '18 at 16:05
• 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. – user May 8 '18 at 17:04
• 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. – Xander Henderson May 8 '18 at 22:39

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 $\pi/2$.

• Most complete answer, IMHO. – user May 8 '18 at 15:40

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

...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:

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.

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.

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.

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