English

Evaluate the following limit : limx→7[(x3-73)(x3+73)x-7]

Advertisements
Advertisements

Question

Evaluate the following limit :

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`

Sum
Advertisements

Solution

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`

= `lim_(x -> 7) ((root(3)(x))^2 - (root(3)(7))^2)/(x - 7)`

= `lim_(x -> 7) (x^(2/3) - 7^(2/3))/(x - 7)`

= `2/3(7)^(-1/3)      ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`

= `2/(3(root(3)(7))`.

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:

If `lim_(x -> 1)[(x^4 - 1)/(x - 1)]` = `lim_(x -> "a")[(x^3 - "a"^3)/(x - "a")]`, find all possible values of a


Evaluate the following limit :

`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following :

`lim_(x -> 0)[x/(|x| + x^2)]`


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 -> - 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 -> 0) sec x`


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):}`


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 ->) (x^"m" - 1)/(x^"n" - 1)`, m and n are integers


Evaluate the following limits:

`lim_(x -> 1) (sqrt(x) - x^2)/(1 - sqrt(x))`


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 -> 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")`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`


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 -> 0) (sin("a" + x) - sin("a" - x))/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 -> pi) (sin3x)/(sin2x)`


Evaluate the following limits:

`lim_(x -> pi) (1 + sinx)^(2"cosec"x)`


Evaluate the following limits:

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


Choose the correct alternative:

`lim_(x - pi/2) (2x - pi)/cos x`


Choose the correct alternative:

`lim_(x -> 0) (8^x - 4x - 2^x + 1^x)/x^2` =


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


If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.


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


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


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


If f(x) is differentiable at x = 1 and `lim_(h → 0) 1/h f(1 + h) = 5`, then f' (1) is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×