Question

Discuss the continuity of the following function at the point(s) or in the interval indicated against them.

`f(x) = (3^x + 3^-x - 2)/x^2`  for x ≠ 0.

= (log3)²                         for x = 0 at x = 0

Answer 1

f(0) = (log3)²   ...(given)

`lim_(x→0) "f"(x) = lim_(x→0) (3^x + 3^-x - 2)/x^2`

= `lim_(x→0) (3^x + 1/3^x - 2)/x^2`

= `lim_(x→0) ((3^x)^2 + 1 - 2(3^x))/(x^2 . (3)^x)`

= `lim_(x→0) (3^x - 1)^2/(x^2 . (3)^x)`  ...[∵ a² - 2ab + b² = (a - b)²]

= `lim_(x→0) [((3^x - 1)/x)^2 xx 1/3^x]`

= `lim_(x→0) ((3^x - 1)/x)^2 xx 1/(lim_(x→0)3^x)`

= `(log 3)^2 xx 1/3^0  ....[because  lim_{x→0}(("a"^"n" - 1)/x) = log "a"]`

= `(log 3)^2 xx 1/1`

= `(log 3)^2`

∴ `lim_(x→0) "f"(x) = f(0)`

∴ f is continuous at x = 0