Advertisements
Advertisements
प्रश्न
Evaluate the following limit :
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
उत्तर
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
= `lim_(z -> "a") (("z" + 2)^(3/2) - ("a" + 2)^(3/2))/((z + 2) - ("a" + 2))`
Put z + 2 = y and a + 2 = b
As z → a, z + 2 → a + 2
i.e. y → b
∴ `lim_(z -> "a") ((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")`
= `lim_(y -> "b") (y^(3/2) - "b"^(3/2))/(y - "b")`
= `3/2* "b"^(1/2) ...[because lim_(z -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= `3/2 sqrt("a" + 2)` ...[∵ b = a + 2]
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(z -> -3) [sqrt("z" + 6)/"z"]`
Evaluate the following limit:
`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
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_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 2)]`
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 the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 2) (x^2 - 1)` = 3
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 1) (x^2 + x + 1)` = 3
Evaluate the following :
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`
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 -> - 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 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 -> 3) (4 - x)`
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 -> 0) sec x`
Write a brief description of the meaning of the notation `lim_(x -> 8) f(x)` = 25
Evaluate the following limits:
`lim_(x -> 2) (1/x - 1/2)/(x - 2)`
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) - 1)/x^2`
Evaluate the following limits:
`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`
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`
Evaluate the following limits:
`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`
Evaluate the following limits:
`lim_(x -> 0) (sinalphax)/(sinbetax)`
Evaluate the following limits:
`lim_(x -> 0) (2^x - 3^x)/x`
Evaluate the following limits:
`lim_(x -> 0) (3^x - 1)/(sqrt(x + 1) - 1)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
Choose the correct alternative:
`lim_(x -> 0) sqrt(1 - cos 2x)/x`
Choose the correct alternative:
`lim_(x -> 0) (x"e"^x - sin x)/x` is
Choose the correct alternative:
`lim_(x -> 0) ("e"^(sin x) - 1)/x` =
If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.
`lim_(x -> 5) |x - 5|/(x - 5)` = ______.
`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.