Question

If $f(x)= \frac{1-cos (7(x-\pi))} {5(x-\pi)^{2}}$, for x≠π is continuous at x=π, find f(π).

Answer 1

concept : any function, y = f(x) is continuous at x = a, only when

$\displaystyle\lim_{x\to a}f(x)=f(a)$

here, f(x) = {1 - cos7(x - π)}/{5(x - π)²} question said, function is continuous at x = π. then we have to find value of f(π)

so, f(π) = $\displaystyle\lim_{x\to\pi}\frac{1-cos7(x-\pi)} {5(x-\pi)^2}$

substitute, h = x - π

then, x→ π coverts into h → 0

now, $\displaystyle\lim_{h\to 0}\frac{1-cos7(h)}{5(h)^2}$

we know, 1 - cos2Φ= 2sin²Φ

so, 1 - cos7h = 2sin²(7h/2)

now, $\displaystyle\lim_{h\to 0}\frac{2sin^2(7h/2)} {5h^2}$

= 2 × $\displaystyle\lim_{h\to 0}\frac{sin^2(7h/2)}{\frac{5(7h/2)^2}{(7/2)^2}}$

= 2 × (7/2)²/5 × 1

= 49/10

hence, f(π) = 49/10

Answer 2

Answer:

Step-by-step explanation: