Advertisements
Advertisements
प्रश्न
Evaluate the following limit:
`lim_(z -> -3) [sqrt("z" + 6)/"z"]`
Advertisements
उत्तर
`lim_(z -> -3) sqrt("z" + 6)/"z"`
= `(lim_(z -> - 3) sqrt(z + 6))/(lim_(z -> - 3) "z") ...[because lim_(z -> -3) "z" ≠ 0]`
= `sqrt(-3 + 6)/-3`
= `sqrt(3)/-3`
= `-1/sqrt(3)`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(x -> 3)[sqrt(2x + 6)/x]`
Evaluate the following limit :
`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the following limit :
`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following :
`lim_(x -> 0) [(sqrt(1 - cosx))/x]`
Evaluate the following :
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.2911 | 0.2891 | 0.2886 | 0.2886 | 0.2885 | 0.28631 |
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) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`
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) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Evaluate the following limits:
`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Evaluate the following limits:
`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x - 1) - 2)/(x - 5)`
Evaluate the following limits:
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`
Find the left and right limits of f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 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) (1 + x - 3x^3)/(1 + x^2 +3x^3)`
A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of salt water after t minutes (in grams per litre) is C(t) = `(30"t")/(200 + "t")`. What happens to the concentration as t → ∞?
Evaluate the following limits:
`lim_(x -> oo) (1 + 3/x)^(x + 2)`
Evaluate the following limits:
`lim_(x -> 0) (2 "arc"sinx)/(3x)`
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`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
Choose the correct alternative:
`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =
`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.
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 ______.
`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.
The value of `lim_(x rightarrow 0) (sqrt((1 + x^2)) - sqrt(1 - x^2))/x^2` is ______.
