The first thing to note is the identity

`cos^(-1)x = pi/2 - sin^(-1)x`

so that

`cos^(-1)x - pi/2 = -sin^(-1)x`

We then require

`lim_(x->0) (-sin^(-1)x)/x`

Since the limit has the indeterminate form 0/0 (taking solutions of inverse sine in (`-pi/2,pi/2` ) only), we can use L'Hopital's rule, which says that the limit as x tends to zero of an expression with numerator and denominator that both tend to zero, is the limit of the ratio of the derivative of the numerator to the derivative of the denominator of that expression.

Here the numerator is `-sin^(-1)x` and the denominator is `x` . Now, since

`d/dx (-sin^(-1)x) = -1/sqrt(1-x^2)` and `d/dx x = 1`

then we have from L'Hopital's rule that

`lim_(x->0) (-sin^(-1)x)/x = lim_(x->0) -1/sqrt(1-x^2) = -1`

**The limit of the expression as x tends to zero is -1.**