Question

If f(x) = `("e"^(2"x") - 1)/"ax"`   , for x < 0 , a ≠ 0

         = 1                                    for x = 0

       = `("log" (1 + 7"x"))/"bx"` , for x > 0 , b ≠ 0

is continuous at x = 0, then find a and b.

Answer 1

Given function is continuous at x = 0. 

`therefore lim_(x->0^-) "f"("x") = lim_(x->0^+) "f(x)" = "f"(0)`

f(0) = 1

`lim_(x->0^-) "f(x)" = lim_(x->0) ("e"^("2x")-1)/"ax"` 

                   `[because lim_(x->0) ("a"^"x" - 1)/"x" = "log a"]`

`= 1/ "a"  lim_(x->0)  ("e"^"2x" - 1)/"2x" xx 2`

`= 1/"a"  "log e" xx 2`

`= 2/"a"`

`therefore 2/"a" = 1 => a = 2`

`lim_(x->1^+) "f(x)" = lim_(x->0) ("log"(1 + 7"x"))/"bx"`

`= 1/"b" lim_(x->0) "log" (1 + 7"x")^(1/"x")`

`= 1/"b"  "log" [lim_(x->0) (1 + 7"x")^(1/"7x")]^7`

`= 1/"b"  "log"  "e"^7 = 1/"b" . 7  "log e"`

`= 7/"b"`

`therefore 7/"b" = 1 => b = 7`