How to evaluate $\lim_{x\rightarrow 0^-} \sec^{-1}(1+x)$?

Question

How to do the following $$\lim_{x\rightarrow 0^-} \sec^{-1}(1+x)$$

I am confused that it seems now $(1+x)<1$ when $x\rightarrow 0^-$, which is not valid for the domain of $\sec^{-1}$. Can anyone give me kick on this?

Answer 1

Note that $$\sec^{-1}y$$

is definded for $$y\in(-\infty,-1]\cup[1,+\infty) $$

thus the limit for $x\to0^-$ that is for $(1+x)\to 1^-$ is not defined.

For continuity

$$\lim_{x\rightarrow 0^+} \sec^{-1}(1+x)=\sec^{-1}1=0$$