Question

If `"f"("x") = ("sin" ("e"^("x"-2) - 1))/("log" ("x" - 1)), "x" ne 2 and "f" ("x") = "k"` for x = 2, then value of k for which f is continuous is ____________.

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Answer 1

If `"f"("x") = ("sin" ("e"^("x"-2) - 1))/("log" ("x" - 1)), "x" ne 2 and "f" ("x") = "k"` for x = 2, then value of k for which f is continuous is 1.

Explanation:

as `lim_("x" -> 2) ("sin" ("e"^("x" - 2) - 1))/("log" ("x" - 1)) = lim_("h" -> 0) ("sin" ("e"^"h" - 1))/("log" (1 + "h"))`

On substituting h = x - 2

`= lim_("h" -> 0) ("sin" ("e"^"h" - 1))/("e"^"h" - 1). ("e"^"h" - 1)/"h" . "h"/("log"(1 + "h"))`

`= 1 . 1 . 1`

`= 1 and "f" (2) = "k"`