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Evaluate the following : limx→1[abx-axbx2-1] - Mathematics and Statistics

Sum

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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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 7 Limits
Miscellaneous Exercise 7 | Q II. (13) | Page 159
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