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 -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
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) [(sqrt(1 - cosx))/x]`
In problems 1 – 6, using the table estimate the value of the limit.
`lim_(x -> 2) (x - 2)/(x^2 - x - 2)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.344820 | 0.33444 | 0.33344 | 0.333222 | 0.33222 | 0.332258 |
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):}`
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 -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`
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 + "k"/x)^("m"/x)`
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 - oo){x[log(x + "a") - log(x)]}`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(2) - sqrt(1 + cosx))/(sin^2x)`
Evaluate the following limits:
`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Choose the correct alternative:
`lim_(theta -> 0) (sinsqrt(theta))/(sqrt(sin theta)`
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.
