Advertisements
Advertisements
प्रश्न
Examine the following function for continuity at the indicated point.
f(x) = `{((x^2 - 4)/(x-2) "," if x ≠ 2),(0 "," if x = 2):}` at x = 2
उत्तर
f(x) = `(x^2 - 4)/(x-2)`, also given that f(2) = 0
`"L"[f(x)]_(x=2) = lim_(x->2^-) f(x)`
[∵ x = 2 – h, where h → 0, x → 2]
`= lim_(h->0)` f(2 - "h") ..[∵ x = 2]
`= lim_(h->0) ((2 - "h")^2 - 4)/((2-"h") - 2)`
`= lim_(h->0) (4 + "h"^2 - 4"h" - 4)/(2 - "h" - 2)`
`= lim_(h->0) ("h"^2 - 4"h")/(-"h")`
`= lim_(h->0) ("h"("h - 4"))/(- "h")`
`= lim_(h->0)` h - 4
`= lim_(h->0) (0 - 4)/(-1)` = 4
But `"L"[f(x)]_(x=2)` f(2) = 0
∴ `"L"[f(x)]_(x=2) ne` f(2)
∴ The given function is not continuous at x = 2.
APPEARS IN
संबंधित प्रश्न
Evaluate the following:
`lim_(x->a) (x^(5/8) - a^(5/8))/(x^(2/3) - a^(2/3))`
If f(x) = `(x^7 - 128)/(x^5 - 32)`, then find `lim_(x-> 2)` f(x)
Find the derivative of the following function from the first principle.
log(x + 1)
Show that the function f(x) = 2x - |x| is continuous at x = 0
If f(x) = `{(x^2 - 4x if x >= 2),(x+2 if x < 2):}`, then f(0) is
`lim_(theta->0) (tan theta)/theta` =
`"d"/"dx" (1/x)` is equal to:
If y = e2x then `("d"^2"y")/"dx"^2` at x = 0 is:
If y = log x then y2 =
`"d"/"dx" ("a"^x)` =