find the limit

Question

lim   f(x)=0

x→1

 

lim  f'(x)=π²/2

x→1

 

lim  f"(x)=a

x→1

 

lim  (x-1)(f(x))÷1-sinπ/2x

x→1

Answer 1

The problem is to find the limit as x→1 of

 

[(x-1)f(x)] / [1-sin(πx/2)]

 

Since the limit is an indeterminate form, use L'Hopital's rule by evaluating the limit of the ratio of the derivatives.

 

d/dx [(x-1)f(x)] = f(x) + (x-1)f'(x)

 

d/dx [1-sin(πx/2)] = -(π/2)cos(πx/2)

 

Limit as x→1 of [f(x) + (x-1)f'(x)] / [-(π/2)cos(πx/2)] is 

 

[f(1) + (1-1)f'(1)] / [-(π/2)cos(π/2)] = [0 + (0)(π²/2)] / [-(π/2)(0)] → 0/0

 

This is still indeterminate.  Applying L'Hopital's rule again, the form becomes

 

[f'(x) + f'(x) + (x-1)f''(x)] / [(π/2)²sin(πx/2)] = [2f'(x) + (x-1)f''(x)] / [(π/2)²sin(πx/2)]

 

As x→1, we get

 

[2f'(1) + (0)f''(1)] / [(π/2)²sin(π/2)] = [2(π²/2) + (0)(a)] / [(π/2)²(1)] = 2(π²/2) / (π/2)²

 

= π² / (π²/4) = 4