Question

If f(x) = (1 - sinx)/(π - 2)^2 , for x ≠ π/2 is continuous at x = π/2, find f(π/2).

Answer 1

if f(x) = (1 - sinx)/(π - 2x)² , for x ≠ π/2 is continuous at x = π/2. we have to find f(π/2)

we know, y = f(x) is continuous at x = a when $\displaystyle\lim_{x\to a}f(x)=f(a)$

so, f(π/2) = $\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{(1-sinx)}{(\pi -2x)^2}$

form of the limit is 0/0

substitute, x = h + π/2

so, f(π/2) = $\displaystyle\lim_{h+\pi/2\to\pi/2}\frac{1-sin(h+\pi/2)}{(\pi-2(h+\pi/2))^2}$

= $\displaystyle\lim_{h\to 0}\frac{1-cosh}{(-2h)^2}$

we know, 1 - cos2Φ = 2sin²Φ

so, 1 - cosh = 2sin²(h/2)

= $\displaystyle\lim_{h\to 0}\frac{2sin^2(h/2)}{4h^2}$

= $\frac{1}{2}\displaystyle\lim_{h\to 0}\frac{sin^2(h/2)}{(h/2)^2\times 4}$

= $\frac{1}{2}\times\frac{1}{4}$

= 1/8

hence, f(π/2) = 1/8