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Question
Evaluate the following :
`lim_(x -> 1) [("ab"^x - "a"^x"b")/(x^2 - 1)]`
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Solution
`lim_(x -> 1) ("ab"^x - "a"^x"b")/(x^2 - 1)`
= `lim_(x -> 1) ("ab"("b"^(x - 1) - "a"^(x - 1)))/(x^2 - 1^2)`
= `lim_(x -> 1) ("ab"("b"^(x - 1) - "a"^(x - 1)))/((x - 1)(x + 1))`
Put x = 1 + h, ∴ x – 1 = h
As x → 1, h → 0
∴ `lim_(x -> 1) ("ab"^x - "a"^x"b")/(x^2 - 1)`
= `lim_("h" -> 0) ("ab"("b"^"h" - "a"^"h"))/("h"(1 + "h" + 1))`
= `"ab" lim_("h" -> 0) ("b"^"h" - 1 + 1 - "a"^"h")/("h"(2 + "h"))`
= `"ab" lim_("h" -> 0) (("b"^"h" - 1) - ("a"^"h" - 1))/("h"(2 + "h"))`
= `"ab" lim_("h" -> 0) 1/(2 + "h") (("b"^"h" - 1)/"h" - ("a"^"h" - 1)/"h")`
= `"ab"* 1/(lim_("h" -> 0)(2 + "h")) (lim_("h" -> 0) ("b"^"h" - 1)/"h" - lim_("h" -> 0) ("a"^"h" - 1)/"h")`
= `"ab"* 1/(2 + 0) * (log"b" - log"a") ...[because lim_(x -> 0) ("a"^x - 1)/x = log"a"]`
= `"ab"/2 log ("b"/"a")`
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