मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

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

Advertisements
Advertisements

प्रश्न

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
तक्ता
Advertisements

उत्तर

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९५]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 4 | पृष्ठ ९५

संबंधित प्रश्‍न

Evaluate the following limit :

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


Evaluate the following limit :

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


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`


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_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`


Evaluate the following limits:

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


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) (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) (sinalphax)/(sinbetax)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

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


Choose the correct alternative:

If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p 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 -> 0) ((2 + x)^5 - 2)/((2 + x)^3 - 2)` = ______.


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


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×