English

Evaluate the following limit : limx→7[x3-343x-7]

Advertisements
Advertisements

Question

Evaluate the following limit :

`lim_(x -> 7) [(x^3 - 343)/(sqrt(x) - sqrt(7))]`

Sum
Advertisements

Solution

`lim_(x -> 7) [(x^3 - 343)/(sqrt(x) - sqrt(7))]`

= `lim_(x -> 7)[(((x^3 - 343)/(x - 7)))/(((sqrt(x) - sqrt(7))/(x - 7)))]    ...[(because x -> 7","  x ≠ 7),(therefore x - 7 ≠ 0)]`

= `(lim_(x -> 7) ((x^3 - 7^3)/(x - 7)))/(lim_(x -> 7) ((x^(1/2) - 7^(1/2))/(x - 7))`

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

= `6*(7)^(5/2)`

= `6 xx 7^2 xx sqrt(7)`

= `294sqrt(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:

`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 -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 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)(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)[x/(|x| + x^2)]`


Evaluate the following :

`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`


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

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 -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`


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 -> 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 -> 1) sin pi x`


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


Evaluate the following limits:

`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`


Evaluate the following limits:

`lim_(x -> 0) (sin^3(x/2))/x^2`


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) (1 - cos^2x)/(x sin2x)`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`


Evaluate the following limits:

`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`


Evaluate the following limits:

`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`


Evaluate the following limits:

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


Choose the correct alternative:

`lim_(x -> oo) sinx/x`


Choose the correct alternative:

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


Choose the correct alternative:

`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =


Choose the correct alternative:

`lim_(x -> 0) ("a"^x - "b"^x)/x` =


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)` =


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) (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 ______.


`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×