मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Evaluate the following limits: limx→22-x+223-4-x3 - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following limits:

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

बेरीज
Advertisements

उत्तर

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

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

= `lim_(x -> 2) ((x + 2)^(1/2) - (2)^(1/2))/(x - 2) xx (x - 2)/((4 - x)^(1/3) - (2)^(1/3))`

= `lim_(x -> 2) ((x + 2)^(1/2) - (4)^(1/2))/((x + 2) - 4) xx (-[(4 - x) - 2])/((4 - x)^(1/3) - (2)^(1/3)]`

= `lim_(x -> 2) ((x + 2)^(1/2)- (4)^(1/2))/((x + 2) - 4) xx - 1/(lim_(x -> 2) ((4 - x)^(1/3) - (2)^(1/3))/((4 - x) - 2)`

`lim_(x -> "a") (x^"n" - "a"^"n") = "na"^("n" - 1)`

= `1/2(4)^(1/2 - 1) xx - 1/(1/3 (2)^(1/3 - 1)`

= `1/2(4)^(-1/2) xx - 3/((2)^(-2/3)`

= `1/(2(2^2)^(1/2)) xx - 3 xx 2^(2/3)`

= `- 1/(2 xx 2) xx 3 xx 2^(2/3)`

= ` - 3/4 xx (2^2)^(1/3)`

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.2 [पृष्ठ १०३]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.2 | Q 11 | पृष्ठ १०३

संबंधित प्रश्‍न

Evaluate the following limit:

`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`


Evaluate the following limit:

`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`


Evaluate the following limit:

`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


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 -> 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


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 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 exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

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


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


Evaluate the following limits:

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


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) (x^3 + x)/(x^4 - 3x^2 + 1)`


Show that `lim_("n" -> oo) (1 + 2 + 3 + ... + "n")/(3"n"^2 + 7n" + 2) = 1/6`


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

`lim_(theta -> 0) (sinsqrt(theta))/(sqrt(sin theta)`


Choose the correct alternative:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×