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

RELATED QUESTIONS

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_(z -> -5)[((1/z + 1/5))/(z + 5)]`


Evaluate the following limit:

`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Evaluate the following limit :

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

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) 1/(x - 3)`


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)`


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`


Sketch the graph of a function f that satisfies the given value:

f(– 2) = 0

f(2) = 0

`lim_(x -> 2) f(x)` = 0

`lim_(x -> 2) f(x)` does not exist.


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


Evaluate : `lim_(x -> 3) (x^2 - 9)/(x - 3)` if it exists by finding `f(3^-)` and `f(3^+)`


Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",",  "for"  x ≠ 1),(0",",  "for"  x = 1):}`


Evaluate the following limits:

`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`


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^3 + x)/(x^4 - 3x^2 + 1)`


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 -> 0) (tan 2x)/x`


Evaluate the following limits:

`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (tan x - sin x)/x^3`


Choose the correct alternative:

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


Choose the correct alternative:

`lim_(x -> 0) ("e"^(sin x) - 1)/x` =


`lim_(x -> 0) ((2 + x)^5 - 2)/((2 + x)^3 - 2)` = ______.


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


`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.


The value of `lim_(x rightarrow 0) (sqrt((1 + x^2)) - sqrt(1 - x^2))/x^2` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×