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


Evaluate the following :

Find the limit of the function, if it exists, at x = 1

f(x) = `{(7 - 4x, "for", x < 1),(x^2 + 2, "for", 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 -> 0) sec x`


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

f(– 2) = 0

f(2) = 0

`lim_(x -> 2) f(x)` = 0

`lim_(x -> 2) f(x)` does not exist.


Evaluate the following limits:

`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`


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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (tan 2x)/(sin 5x)`


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) (2 "arc"sinx)/(3x)`


Evaluate the following limits:

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


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) (1 + sinx)^(2"cosec"x)`


Evaluate the following limits:

`lim_(x -> ) (sinx(1 - cosx))/x^3`


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


The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×