Question

f(x) = `{{:((1 - cos 2x)/x^2",", "if"  x ≠ 0),(5",", "if"  x = 0):}` at x = 0

Answer 1

We have, `{{:((1 - cos 2x)/x^2",", "if"  x ≠ 0),(5",", "if"  x = 0):}` at x = 0

At x = 0

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

= `lim_("h" -> 0) (1 - cos2(0 - "h"))/(0 - "h")^2`

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

= `lim_("h" -> 0) (2 sin^2"h")/"h"^2` = 2

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

= `lim("h" -> 0) (1 - cos2(0 + "h"))/(0 + "h")^2`

= `lim_("h" -> 0) (2sin^2"h")/"h"^2` = 2

Also f(0) = 5 ... (Given)

Since, L.H.L. = R.H.L. ≠ f(0)

Hence, f(x) is discontinuous at x = 0.