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_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> -3) (3x + 2)` = – 7
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (cos x - 1)/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.0001 | 0.01 | 0.1 |
| f(x) | 0.04995 | 0.0049999 | 0.0004999 | – 0.0004999 | – 0.004999 | – 0.04995 |
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 -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`
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):}`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Evaluate the following limits:
`lim_(x -> oo) (1 + x - 3x^3)/(1 + x^2 +3x^3)`
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 -> oo)(1 + 1/x)^(7x)`
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`
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`
Evaluate the following limits:
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Choose the correct alternative:
`lim_(x -> oo) sinx/x`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.
