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Test the continuity of the following function at the point or interval indicated against them : f(x) =x-1-(x-1)13x-2, for x≠2=15, for x=2}at x = 2 - Mathematics and Statistics

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Question

Test the continuity of the following function at the point or interval indicated against them :

f(x)  `{:(= (sqrt(x - 1) - (x - 1)^(1/3))/(x - 2)",",  "for"  x ≠ 2),(= 1/5",",  "for"  x = 2):}}`at x = 2

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Solution

f(2) = `1/5`   ...(Given)  ...(1)

`lim_(x -> 2) "f"(x) =  lim_(x -> 2) (sqrt(x - 1) - (x - 1)^(1/3))/(x - 2)`

Put x = 1 + h. Then as x → 2, h → 1

Also x – 1 = h and x – 2 = (1 + h) – 2 = h – 1

∴ `lim_(x -> 2) "f"(x) =  lim_("h" -> 1) (sqrt("h") - "h"^(1/3))/("h" - 1)`

= `lim_("h" -> 1) ((sqrt("h") - 1) - ("h"^(1/3) - 1))/("h" - 1)`

= `lim_("h" -> 1) [(sqrt("h") - 1)/("h" - 1) - ("h"^(1/3) - 1)/("h" - 1)]`

= `lim_("h" -> 1) (("h"^(1/2) - 1^(1/2))/("h" - 1)) - lim_("h" -> 1) (("h"^(1/3) - 1^(1/3))/("h" - 1))`

= `1/2(1) ^(-1/2) - 1/3(1)^(-2/3)    ...[because  lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`

= `1/2 - 1/3`

= `1/6`

From (1) and (2),

`lim_(x -> 2) "f"(x) ≠ "f"(2)`

∴ f is discontinuous at x = 2

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Continuous and Discontinuous Functions
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Chapter 8: Continuity - EXERCISE 8.1 [Page 173]

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