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Question
Evaluate the following :
`lim_(x -> 2) [(logx - log2)/(x - 2)]`
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Solution
Put x = 2 + h. Then x – 2 = h and as x → 2, h → 0.
∴ `lim_(x -> 2) [(log x - log 2)/(x - 2)]`
= `lim_("h" -> 0) [(log(2 + "h") - log2)/"h"]`
= `lim_("h" -> 0) (log ((2 + "h")/2))/"h"`
= `lim_("h" -> 0) (log(1 + "h"/2))/(("h"/2) xx 2)`
= `1/2 lim_("h" -> 0) (log(1 + "h"/2))/(("h"/2))`
= `1/2 xx 1 ...[because "h" -> 0, "h"/2 -> 0 "and" lim_(x -> 0) (log(1 + x))/x = 1]`
= `1/2`
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