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 $?$


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.


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.

  • $\begingroup$ I'd say the domain of $\sin^{-1} (\sec x) $ is the set $\{n\pi\mid n\in\mathbb N\},$ whose image under $\sec$ is $\{-1,1\}.$ $\endgroup$ – David K Jul 22 at 2:33
  • $\begingroup$ Thank you that's more accurate. $\endgroup$ – zjm Jul 22 at 2:41

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