Question

f(x) = `{{:((sqrt(1 + "k"x) - sqrt(1 - "k"x))/x",",  "if" -1 ≤ x < 0),((2x + 1)/(x - 1)",",  "if"  0 ≤ x ≤ 1):}` at x = 0

Answer 1

We have f(x) = `{{:((sqrt(1 + "k"x) - sqrt(1 - "k"x))/x",",  "if" -1 ≤ x < 0),((2x + 1)/(x - 1)",",  "if"  0 ≤ x ≤ 1):}`

L.H.L. = `lim_(x -> 0^-) (sqrt(1 + "k"x) - sqrt(1 - "k"x))/x`

= `lim_(x -> 0^-) ((sqrt(1 + "k"x) - sqrt(1 - "k"x))/x) * ((sqrt(1 + "k"x) + sqrt(1 - "k"x))/(sqrt(1 + "k"x) + sqrt(1 - "k"x)))`

= `lim_(x -> 0^-) (1 + "k"x - 1 + "k"x)/(x[sqrt(1 + "k"x) + sqrt(1 + "k"x)])`

= `lim_("h" -> 0) (2"k")/(x[sqrt(1 + "k"(0 - "h")) + sqrt(1 - "k"(0 - "h")]`

= `lim_("h" -> 0) (2"k")/(sqrt(1 - "kh") + sqrt(1 + "kh")`

= `(2"k")/2`

= k

R.H.L. = `lim_(x -> 0^+) (2x + 1)/(x - 1)`

= `lim_("h" -> 0) (2(0 + "h") + 1)/((0 + "h") - 1)`

= `lim_("h" -> 0) (2"h" + 1)/("h" - 1)`

= – 1

Also f(0) = `(2 xx 0 + 1)/(0 - 1)` = – 1

We must have L.H.L. = R.H.L. = f(0)

⇒ k = – 1