English

Evaluate the following limit : limy→1[2y-27+y3-2] - Mathematics and Statistics

Advertisements
Advertisements

Question

Evaluate the following limit :

`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 2)]`

Sum
Advertisements

Solution

`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y )- 2)]`

= `lim_(y -> 1) [(2y - 2)/((7 + y)^(1/3) - 8^(1/3))] xx [((7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3))/((7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3))]`

= `lim_(y -> 1) ((2y - 2)[(7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3)])/((7 + y) - 8)   ...[because "a" - "b" = ("a"^(1/3) - "b"^(1/3)) ("a"^(2/3) + "a"^(1/3)"b"^(1/3) + "b"^(2/3))]`

= `lim_(y -> 1) (2(y - 1)[(7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3)])/(y - 1)`

= `lim_(y -> 1) 2[(7 + y)^(2/3) + (7 + y)^(1/3) 8^(1/3) + 8^(2/3)]   ...[(because y -> 1","  y ≠ 1","),(therefore y - 1 ≠ 0)]`

= `2[(7 + 1)^(2/3) + (7 + 1)^(1/3) 8^(1/3) + 8^(2/3)]`

= = 2[4 + 4 + 4] ...`[∵ 8^(1/3) = 2 and 8^(2/3) = 4]`

= 24

shaalaa.com
Concept of Limits
  Is there an error in this question or solution?
Chapter 7: Limits - Exercise 7.1 [Page 139]

RELATED QUESTIONS

Evaluate the following limit:

`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`


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_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


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 -> 1) (x^2 + x + 1)` = 3


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 - 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) (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) (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

Write a brief description of the meaning of the notation `lim_(x -> 8) f(x)` = 25


Evaluate the following limits:

`lim_(sqrt(x) -> 3) (x^2 - 81)/(sqrt(x) - 3)`


Evaluate the following limits:

`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`


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`


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))`


Find the left and right limits of f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 2


Show that `lim_("n" -> oo) (1 + 2 + 3 + ... + "n")/(3"n"^2 + 7n" + 2) = 1/6`


Evaluate the following limits:

`lim_(x -> oo)(1 + "k"/x)^("m"/x)`


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 -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)]`


Evaluate the following limits:

`lim_(x - oo){x[log(x + "a") - log(x)]}`


Evaluate the following limits:

`lim_(x -> pi) (sin3x)/(sin2x)`


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:

If `f(x) = x(- 1)^([1/x])`, x ≤ 0, then the value of `lim_(x -> 0) f(x)` is equal to


Choose the correct alternative:

`lim_(x -> 3) [x]` =


Choose the correct alternative:

If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is


`lim_(x -> 5) |x - 5|/(x - 5)` = ______.


`lim_(x -> 0) (sin 4x + sin 2x)/(sin5x - sin3x)` = ______.


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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×