हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Evaluate the following limits: limx→∞(x2-2x+1x2-4x+2)x

Advertisements
Advertisements

प्रश्न

Evaluate the following limits:

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

योग
Advertisements

उत्तर

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

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

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

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

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

`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 - 4x + 2))^x = [lim_(x -> oo) "e"]^((2x^2 - x)/(x^2 - 4x + 2))`

`lim_(x -> oo) (1 + 1/x)^x`  = e

= `"e"^(lim_(x ->oo)) ((2x^2 - x)/(x^2 - 4x + 2))`

= `"e"^(lim_(x -> oo) (x^2(2 -x/x^2))/(x^2(1 - (4x)/(x^2) + 2/x^2))`

= `"e"^(lim_(x ->oo) ((2 - 1/x)/(1 -4/x + 2/x^2))`

= `"e"^(((2 - 0)/(1 - 0 + 0))`

`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x` = e2 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [पृष्ठ ११८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 24 | पृष्ठ ११८

संबंधित प्रश्न

Evaluate the following limit:

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


Evaluate the following limit:

`lim_(x -> 3)[sqrt(2x + 6)/x]`


Evaluate the following limit :

`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


Evaluate the following limit :

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


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)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


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


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

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


Choose the correct alternative:

`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =


If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×