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Question
`f(x) = (log x - log 3)/(x - 3)` for x ≠ 3
= 3 for x = 3, at x = 3.
Solution
f(3) = 3 …[given]
`lim_(x→3) "f"(x) = lim_(x→3) (log x - log 3)/(x - 3)`
Substitute x – 3 = h
∴ x = 3 + h,
as x → 3, h → 0
∴ `lim_(x→3) "f"(x) = lim_("h"→ 0) (log("h" + 3) - log 3)/(3 + "h" - 3)`
= `lim_("h"→ 0) log(("h" + 3)/3)/"h"`
= `lim_("h" → 0) (log(1 + "h"/3))/(("h"/3))xx 1/3`
= `1/3 lim_("h"→ 0} (log(1 + "h"/3))/(("h"/3))`
= `1/3(1)` ...`[∵ lim_(x → 0) log(1 + x)/x = 1]`
= `1/3`
∴ `lim_(x → 3} "f"(x) ≠ "f"(3)`
∴ f is discontinuous at x = 3
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