हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा 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))`

योग

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
Concept of Limits
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 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
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.2 | Q 11 | पृष्ठ १०३

संबंधित प्रश्न

Evaluate the following limit:

`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`


Evaluate the following limit:

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


Evaluate the following :

`lim_(x -> 0)[x/(|x| + x^2)]`


Evaluate the following :

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


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)]}`


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 -> - 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 -> 5) |x - 5|/(x - 5)`


Write a brief description of the meaning of the notation `lim_(x -> 8) f(x)` = 25


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x  -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

`lim_(x -> 0) ("e"^(sin x) - 1)/x` =


Choose the correct alternative:

The value of `lim_(x -> 0) sinx/sqrt(x^2)` is


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×