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Find f(a), if f is continuous at x = a where, f(x) = 1-cos[7(x-π)]5(x-π)2, for x ≠ π at a = π

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Question

Find f(a), if f is continuous at x = a where,

f(x) = `(1 - cos[7(x - pi)])/(5(x - pi)^2`, for x ≠ π at a = π

Sum
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Solution

f is continuous at x = π

∴ f(π) = `lim_(x -> pi) "f"(x) =  lim_(x -> pi) (1 - cos[7 (x - pi)])/(5(x - pi)^2`

Put x – π = h, as x → π, h → 0

∴ f(π) = `lim_("h" -> 0) (1 - cos7"h")/(5"h"^2)`

= `lim_("h" -> 0) (2sin^2((7"h")/2))/(5"h"^2)`

= `2/5 lim_("h" -> 0) (sin^2 ((7"h")/2))/(((7"h")/2)^2) xx (7/2)^2`

= `2/5 |lim_("h" -> 0) (sin ((7"h")/2))/(((7"h")/2))|^2 xx 49/4`

= `2/5 xx (1)^2 xx 49/4  ...[because lim_(theta -> 0) sintheta/theta = 1]`

∴ f(π) = `49/10`

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Chapter 8: Continuity - MISCELLANEOUS EXERCISE-8 [Page 178]

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