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 -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
Evaluate the following limit :
`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`
Evaluate the following limit :
If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate the following limit :
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 2)(2x + 3)` = 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 -> 2) (x - 2)/(x^2 - 4)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.25641 | 0.25062 | 0.250062 | 0.24993 | 0.24937 | 0.24390 |
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) sin x/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.99833 | 0.99998 | 0.99999 | 0.99999 | 0.99998 | 0.99833 |
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) (x^2 + 2)`
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) - 1)/x^2`
Evaluate the following limits:
`lim_(x -> 0)(1 + x)^(1/(3x))`
Evaluate the following limits:
`lim_(x -> oo) (1 + 3/x)^(x + 2)`
Evaluate the following limits:
`lim_(x -> 0) (sin^3(x/2))/x^2`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/(sin 5x)`
Evaluate the following limits:
`lim_(x -> 0) (tan x - sin x)/x^3`
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
`lim_(x -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is
Choose the correct alternative:
The value of `lim_(x -> 0) sinx/sqrt(x^2)` is
If `lim_(x -> 1) (x + x^2 + x^3|+ .... + x^n - n)/(x - 1)` = 820, (n ∈ N) then the value of n is equal to ______.
