Question

If f(x) = `(e^(2x) - 1)/(ax)` .                for x < 0 , a ≠ 0
         = 1.                             for x = 0
         = `(log(1 + 7x))/(bx)`.        for x > 0 , b ≠ 0
is continuous at x = 0 . then find a and b

Answer 1

Since f(x) is continuous  at x = 0 

`lim_(x→0^-) f(x) = lim_(x→0^+) f(x) = f(0)`

Now `lim_(x→0^-) f(x) = lim_(x→0) (e^(2x) - 1)/(ax)`

= `2/a[lim_(x→0) (e^(2x) - 1)/(2x)]`

= `2/a[log e]`   [∵ `lim_(x→0) (a^x - 1)/x = log a`]

= `2/a[1] = 2/a`

`lim_(x→0^-) f(x) = 2/a`

Now `lim_(x→0^+) f(x) = lim_(x→0) log(1 + 7x)/(bx)`

= `7/b[lim_(x →0) log(1 + 7x)/(7x)]`

= `7/b[1]        [∵ lim_(x→0) log(1 + x)/x = 1]`

∴ `lim_(x→0^+) f(x) = 7/b`

Also f(0) = -1

∴ `lim_(x→0^-) f(x) = lim_(x→0^+) f(x) = f(0)`

∴ `2/a = 7/b = 1`

∴ a = 2 , b = 7