English
Tamil Nadu Board of Secondary EducationHSC Science Class 11

In problems 1 – 6, using the table estimate the value of the limitlimx→0x+3-3x 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

Advertisements
Advertisements

Question

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
Chart
Advertisements

Solution

Let f(x) = `lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`

x – 0.1  – 0.01 – 0.001 0.001 0.01 0.1
f(x)

`(sqrt(3 - 0.1) - sqrt(3))/(- 0.1)`

= `(1.703 - 1.732)/(- 0.1)`

= 0.29

`(sqrt(3 - 00.1) - sqrt(3))/(- 0.01)`

= `(1.703 - 1.732)/(- 0.01)`

= 0.288

`(sqrt(3 - 0.001) - sqrt(3))/(- 0.001)`

= `(0.000284)/(- 0.001)`

= 0.288

`(sqrt(3 + 0.001) - sqrt(3))/(- 0.001)`

= `(0.000289)/(0.001)`

= 0.289

`(sqrt(3 + 0.01) - sqrt(3))/(0.01)`

= `(0.00288)/(0.01)`

= 0.288

`(sqrt(3 + 0.1) - sqrt(3))/(0.1)`

= `(0.0286)/(0.1)`

= 0.286

`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x` = 0.288

shaalaa.com
  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 3 | Page 95

RELATED QUESTIONS

Evaluate the following limit :

`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`


Evaluate the following limit :

`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`


Evaluate the following :

Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`


In problems 1 – 6, using the table estimate the value of the limit.

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

x 1.9 1.99 1.999 2.001 2.01 2.1
f(x) 0.344820 0.33444 0.33344 0.333222 0.33222 0.332258

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 -> 1) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`


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

f(0) is undefined

`lim_(x -> 0) f(x)` = 4

f(2) = 6

`lim_(x -> 2) f(x)` = 3


Evaluate the following limits:

`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`


Evaluate the following limits:

`lim_(x -> 5) (sqrt(x - 1) - 2)/(x - 5)`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0)(1 + x)^(1/(3x))`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x`


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

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


Choose the correct alternative:

The value of `lim_(x -> 0) sinx/sqrt(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×