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 -> -3) [sqrt("z" + 6)/"z"]`
Evaluate the following limit:
`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`
Evaluate the following limit :
`lim_(x -> 1) [(x + x^3 + x^5 + ... + x^(2"n" - 1) - "n")/(x - 1)]`
Evaluate the following :
`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`
| x | – 3.1 | – 3.01 | – 3.00 | – 2.999 | – 2.99 | – 2.9 |
| f(x) | – 0.24845 | – 0.24984 | – 0.24998 | – 0.25001 | – 0.25015 | – 0.25158 |
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 -> 1) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`
Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.
f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`
Sketch the graph of a function f that satisfies the given value:
f(0) is undefined
`lim_(x -> 0) f(x)` = 4
f(2) = 6
`lim_(x -> 2) f(x)` = 3
Evaluate the following limits:
`lim_(x -> 2) (x^4 - 16)/(x - 2)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
Evaluate the following limits:
`lim_(x -> oo) (1 + x - 3x^3)/(1 + x^2 +3x^3)`
Show that `lim_("n" -> oo) (1 + 2 + 3 + ... + "n")/(3"n"^2 + 7n" + 2) = 1/6`
Show that `lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = 9/25`
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) (2 "arc"sinx)/(3x)`
Choose the correct alternative:
`lim_(x -> 3) [x]` =
Choose the correct alternative:
`lim_(x -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is
If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.
