English
Tamil Nadu Board of Secondary EducationHSC Science Class 11

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

Advertisements
Advertisements

Question

Evaluate the following limits:

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

Sum

Solution

`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
Concept of Limits
  Is there an error in this question or solution?
Chapter 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [Page 118]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 24 | Page 118

RELATED QUESTIONS

Evaluate the following limit:

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


Evaluate the following limit :

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


In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

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


In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`

x – 0.1  – 0.01 – 0.001 0.001 0.01 0.1
f(x) 0.2911 0.2891 0.2886 0.2886 0.2885 0.28631

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 -> 2) (x^4 - 16)/(x - 2)`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`


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_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`


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


Choose the correct alternative:

`lim_(x -> 3) [x]` =


Choose the correct alternative:

`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×