Question

Find k if the following function is continuous at the points indicated against them:

`f(x) = ((5x - 8)/(8 - 3x))^(3/(2x - 4)`  for x ≠ 2
         = k                           for x = 2 at x = 2.

Answer 1

f is continuous at x = 2

∴ f(2) = `lim_(x→2) "f"(x)`

∴ k = `lim_(x→2) ((5x - 8)/(8 - 3x))^(3/(2x - 4)`

Substitute x - 2 = h
∴ x = 2 + h
As x → 2, h → 0

∴ k = `lim_("h"→0)[(5(2 + "h") - 8)/(8 - 3(2 + "h"))]^(3/(2(2 + "h") - 4)`

= `lim_("h"→0) ((10 + 5"h" - 8)/(8 - 6 - 3"h"))^(3/(2"h")`

= `lim_("h"→0} ((2 + 5"h")/(2 - 3"h"))^(3/(2"h")`

= `lim_("h"→0) [(2(1 + (5"h")/2))/(2(1 - (3"h")/2))]^(3/(2"h")`

= `lim_("h"→0) (1 + (5"h")/2)^(3/(2"h"))/((1 - (3"h")/2)^(3/(2"h")))`

= `(lim_("h"→0) [(1 + (5"h")/2)^(2/(5"h")]]^(5/2 xx 3/2))/(lim_("h"→0)[(1 - (3"h")/2)^((-2)/(3"h"))]^((-3)/2 xx 3/2))`

= `"e"^(15/4)/"e"^((-9)/4)`  ....`[(because "h" → 0","(5"h")/2 → 0"," (-3h)/2 → 0), ("and" lim_(x→0) (1 + x)^(1/x) = "e")]`

= `"e"^(24/4)`

∴ k = e⁶