Question

Examine the following function for continuity at the indicated point.

f(x) = `{((x^2 - 9)/(x-3) "," if  x ≠ 3),(6 "," if x = 3):}` at x = 3

Answer 1

Given that f(x) = `(x^2 - 9)/(x-3)`, and f(3) = 6

`"L"[f(x)]_(x=3) = lim_(x->3^-) f(x)`

`= lim_(h->0)` f(3 - h)   

`= lim_(h->0) ((3 - "h")^2 - 9)/(3-"h" - 3)`

`= lim_(h->0) (9 + "h"^2 - 6"h" - 9)/(- "h")`

`= lim_(h->0) ("h"^2 - 6"h")/(-"h")`

`= lim_(h->0) ("h"("h - 6"))/(- "h")`

`= lim_(h->0) (h - 6)/(-1)`

`= lim_(h->0) (0 - 6)/(-1)` = 6

`"R"[f(x)]_(x=3^+) = lim_(x->3^+) f(x)`

[∵ x = 3 + h, where h → 0, x → 3]

`= lim_(h->0) "f"(3 + "h")`

`= lim_(h->0) ((3 + "h")^2 - 9)/(3 + "h" - 3)`

`= lim_(h->0) (9 + "h"^2 + 6"h" - 9)/"h"`

`= lim_(h->0) ("h"^2 + 6"h")/"h"`

`= lim_(h->0) ("h"("h + 6"))/"h"`

= 0 + 6

= 6

Also given that f(3) = 6

Thus `"L"[f(x)]_(x=3) = "R"[f(x)]_(x=3) = f(3)`

i.e., `lim_(x->3^-) "f"(x) = lim_(x->3^+) "f"(x)` = f(3)

∴ The given function f(x) is continuous at x = 3.