Advertisements
Advertisements
प्रश्न
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Advertisements
उत्तर
No, f(x) = 4, It is the value of the function at x = 2
This limit doesn’t exists at x = 2
Since f(2) = 4
It need not imply that `lim_(x -> 2^-) f(x) = lim_(x -> 2^+) f(x)`
∴ We cannot conclude at x = 2
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit :
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 5) |x - 5|/(x - 5)`
Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.
f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`
Evaluate : `lim_(x -> 3) (x^2 - 9)/(x - 3)` if it exists by finding `f(3^-)` and `f(3^+)`
Evaluate the following limits:
`lim_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`
Evaluate the following limits:
`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`
Find the left and right limits of f(x) = tan x at x = `pi/2`
Evaluate the following limits:
`lim_(x -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`
Evaluate the following limits:
`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`
Evaluate the following limits:
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + "a"^2) - "a")/(sqrt(x^2 + "b"^2) - "b")`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
Choose the correct alternative:
`lim_(x -> 0) (8^x - 4x - 2^x + 1^x)/x^2` =
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.
`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.
