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

Question

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?

Answer 1

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.

Answer 2

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 $\pi/2$.

Answer 3

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

Answer 4

...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.}

Answer 5

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.