मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Evaluate the following limits: lim(x→∞)x3+xx4-3x2+1

Advertisements
Advertisements

प्रश्न

Evaluate the following limits:

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

बेरीज
Advertisements

उत्तर

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

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

= `((1 + 1/oo))/(oo(1 - 1/oo + 1/oo))`

= `(1 +0)/(oo(1 - 0 + 0))`

= `1/oo`

`lim_(x -> oo) (x^3 + x)/(x^4 - 3x^2 + 1)` = 0

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.3 [पृष्ठ १११]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.3 | Q 4 | पृष्ठ १११

संबंधित प्रश्‍न

Evaluate the following limit :

`lim_(x -> 1) [(x + x^3 + x^5 + ... + x^(2"n" - 1) - "n")/(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 -> 1) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", 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 -> 5) |x - 5|/(x - 5)`


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.


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


If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?


Evaluate the following limits:

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


Evaluate the following limits:

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


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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x - oo){x[log(x + "a") - log(x)]}`


Evaluate the following limits:

`lim_(x -> pi) (sin3x)/(sin2x)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

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


Choose the correct alternative:

`lim_(x -> 0) (8^x - 4x - 2^x + 1^x)/x^2` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×