Advertisements
Advertisements
Question
Show that f(x) = |x| is continuous at x = 0.
Advertisements
Solution
Given that f(x) = |x| = `{(x if x >= 2),(- x if x < 0):}`
`"L"["f"(x)]_(x=0) = lim_(x->0^-)`f(x)
[∵ x = 0 – h]
`= lim_(h->0^-) "f"(0 - "h")`
`= lim_(h->0^-) "f"(- "h")`
`= lim_(h->0^-) |- "h"|`
`= lim_(h->0^-) |"h"|`
`= lim_(h->0^-) "h" = 0`
`"R"["f"(x)]_(x=0^+) = lim_(x->0^+)`f(x)
`= lim_(h->0^+) "f"(0 - "h")`
`= lim_(h->0^+) "f"("h")`
`= lim_(h->0^+) |"h"|`
`= lim_(h->0) "h"`
= 0
[∵ |x| = x if x > 0]
Also f(0) = |0| = 0
`lim_(x->0^-) "f"(x) = lim_(x->0^+ "f"(x))` = f(0)
∴ f(x) is continuous at x = 0.
APPEARS IN
RELATED QUESTIONS
Evaluate the following:
`lim_(x->0) (sqrt(1+x) - sqrt(1-x))/x`
If `lim_(x->2) (x^n - 2^n)/(x-2) = 448`, then find the least positive integer n.
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
Find the derivative of the following function from the first principle.
x2
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
Verify the continuity and differentiability of f(x) = `{(1 - x if x < 1),((1 - x)(2 - x) if 1 <= x <= 2),(3 - x if x > 2):}` at x = 1 and x = 2.
`lim_(theta->0) (tan theta)/theta` =
If y = x and z = `1/x` then `"dy"/"dx"` =
`"d"/"dx" ("a"^x)` =
