How do I solve lim x→π/2 (√sinx - 1)/(x-π/2)?

Question

How do I solve lim x→π/2 (√sinx - 1)/(x-π/2)?

Answer 1

In case you don't know L'Hopital's Rule, we can use the definition of the derivative:

 

Recall:  f'(c) = lim_x→c [(f(x) - f(c)) / (x-c)]

 

f(x) = √sinx.  Then f(π/2) = 1

 

So, the given limit can be written as lim_x→π/2 [(f(x) - f(π/2)) / (x - π/2)] = f'(π/2)

 

     f'(x) = (1/2)(sinx)^-1/2(cosx)    So, f'(π/2) = (1/2)(1)(0) = 0

 

Answer 2

Since this is an indeterminate form, you can use L'Hopital's rule.

 

lim_x→π/2 [(√sinx - 1)/(x - π/2)] = lim_x→π/2 [(1/2)(sinx)^-1/2cosx] = (1/2)cos(π/2) / √sin(π/2) = 0