limit of $\sin^{-1} (\sec x) $ as $x$ tends to $0$ What is the limit of $\sin^{-1}  (\sec x) $ as $x$ tends to $0$.
By direct substitution the value of $\sec x $ at $x =0$ is $1$. So limit should be $1$. 
But answer is given that limit doesn't exist. How $?$
 A: First of all, if you do direct substitution, then yes, 
$\sec 0 = 1,$ but you still have to consider the arc sine function.
And $\sin^{-1} (1) = \frac\pi2.$
So the result of direct substition by $x=0$ is $\sin^{-1} (\sec x) = \frac\pi2.$
But the direct substitution method agrees with the limit only if both of the following conditions are met:


*

*The limit exists.

*The function is continuous.


The existence of the limit as $x\to0$ doesn't depend on the value at $x=0.$
But it does depend on the value of the function when $x$ is very close to zero.
And if $x$ is close to zero, but not equal to zero, we have
$$ \sec x > 1 $$
and since $\sin$ can only produce numbers in the range $[-1,1],$
$\sin^{-1} (\sec x)$ is undefined when $\sec x > 1.$
You can't produce a limit when the function is not defined anywhere near the limit point except at the limit point itself.
A: Direct substitution does not always work. It only works when the given function is continuous about $x_0$ (the number x approaches, in this case 0). Remember the idea of limits is that when x gets "close" to $0, f(x)$ ought to get "close" to some number. In the case of this problem, $\sin^{-1}(x)$ is only defined on the domain $[-1, 1]$, but the range of $sec(x)$ is $(-\infty, -1] \cup [1, \infty)$. Thus, the domain of $\sin^{-1}(\sec(x))$ is only a set of discrete points $\{n\pi : n \in \mathbb{Z}\}$. If you try plotting $\sin^{-1}(\sec(x))$ in desmos, maybe this will be more clear. With just these single points, it is impossible to say what the behavior of $\sin^{-1}(\sec(x))$ is like "close to" $x=0$ as the function is not defined close to $x = 0$. Hence why the limit does not exist.
