English
Tamil Nadu Board of Secondary EducationHSC Science Class 11

In problems 1 – 6, using the table estimate the value of the limitlimx→-31-x-2x+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 - Mathematics

Advertisements
Advertisements

Question

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
Chart

Solution

Let f(x) = `lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`

x – 3.1 – 3.01 – 3.00 – 2.999 – 2.99 – 2.9
f(x)

`(sqrt(1 + 3.1) - 2)/(- 3.1 + 3)`

= `(0.0248)/( - 0.1)`

= – 0.248

`(sqrt(1 + 3.01) - 2)/(- 3.01 + 3)`

= `(0.00250)/( - 0.01)`

= – 0.249

`(sqrt(1 + 3) - 2)/(- 3 + 3)`

= `0/0`

= 0

`(sqrt(1 + 2.999) - 2)/(- 2.999 + 3)`

= `(- 0.000250)/(0.001)`

= – 0.25

`(sqrt(1 + 2.99) - 2)/(- 2.99 + 3)`

= `(- 0.00250)/(0.01)`

= – 0.25

`(sqrt(1 + 2.9) - 2)/(- 2.9 + 3)`

= `(- 0.02515)/(0.1)`

= – 0.2515

`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)` = – 0.25

shaalaa.com
Concept of Limits
  Is there an error in this question or solution?
Chapter 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [Page 95]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 4 | Page 95

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)]`


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)]}`


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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1


Evaluate the following limits:

`lim_(x -> 0) (sinalphax)/(sinbetax)`


Choose the correct alternative:

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


Choose the correct alternative:

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


Choose the correct alternative:

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


Choose the correct alternative:

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


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


`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×