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.
Solution
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`
