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Question
Evaluate the following:
`lim_(x→π/6) [(2sinx − 1)/(π − 6x)]`
Solution
Put `π/6 − "h" = x`
∴ h = `π/6 − x`
Then as `x→π/6, "h"→0`
∴ `lim_(x→π/6) [(2sinx − 1)/(π - 6x)]`
= `lim_("h"→0) [(2sin(π/6 − "h") - 1)/(π − 6(π/6 − "h"))]`
= `lim_("h"→0) (2[sin(π/6 − "h") - 1/2])/(6"h")`
= `1/3 lim_("h"→0) (sin(π/6 − "h") − sin π/6)/"h"`
= `1/3 lim_("h"→0) (2cos((π/6 − "h" + π/6)/2).sin((π/6 − "h" − π/6)/2))/"h"`
= `1/3 lim_("h"→0) (2cos(π/6 − "h"/2).sin("−h"/2))/"h"`
= `(−1)/3 lim_("h"→0) cos(π/6 − "h"/2).sin("h"/2)/("h"/2)`
= `(−1)/3 lim_("h"→0) cos(π/6 − "h"/2). lim_("h"→0) sin("h"/2)/("h"/2)`
= `(−1)/3. cos (π/6 − 0).(1) ...[∵ lim_(θ →0) (sin pθ)/(pθ) = 1]`
= `(−1)/3 × sqrt3/2`
= `(−1)/(2sqrt3)`
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