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Question
The following function has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it become continuous :
f(x) `{:(= log_((1 + 3x)) (1 + 5x)",", "for" x > 0),(=(32^x - 1)/(8^x - 1)",", "for" x < 0):}}` at x = 0
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Solution
`lim_(x -> 0) "f"(x) = lim_(x -> 0) log_((1 + 3x)) (1 + 5x)`
= `lim_(x -> 0) [(log (1 + 5x))/(log(1 + 3x))] ...[because log_"a"x = logx/log"a"]`
= `lim_(x -> 0) ([(log(1 + 5x))/(5x)*5])/([(log (1 + 3x))/(3x)* 3])` ...[∵ x → 0, ∴ x ≠ 0]
= `5/3 (lim_(x -> 0) [(log(1 + 5x))/(5x)])/(lim_(x -> 0) [(log(1 + 3x))/(3x)])`
= `5/3 xx 1/1 ...[because x -> 0, therefore 3x -> 0, 5x -> 0 "and" lim_(x -> 0) (log(1 + x))/x = 1]`
= `5/3` ...(1)
`lim_(x -> 0^-) "f"(x) = lim_(x -> 0) (32^x - 1)/(8^x - 1)`
= `lim_(x -> 0) (((32^x - 1)/x))/(((8^x - 1)/x))` ...[∵ x → 0, ∴ x ≠ 0]
= `(lim_(x -> 0) ((32^x - 1)/x))/(lim_(x -> 0) ((8^x - 1)/x))`
= `log32/log8 ...[because lim_(x -> 0) ("a"^x - 1)/x = log "a"]`
= `(log2^5)/(log2^3)`
= `(5log2)/(3log2)`
= `5/3` ...(2)
From (1) and (2),
`lim_(x -> 0^+) "f"(x) = lim_(x -> 0^-) "f"(x) = 5/3`
∴ `lim_(x -> 0) "f"(x) = 5/3`
f(0) is not defined
∴ f is discontinuous at x = 0.
This discontinuity is removable.
We redefine the function as follows to make it continuous at x = 0
f(x) `{:(= log_((1 + 3x)) (1 + 5x),"," "for" x > 0),(= 5/3,"," "for" x = 0),(=(32^x - 1)/(8^x - 1),"," "for" x < 0):}`
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