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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
Solution
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
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