मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Evaluate the following : limx→0{1x12[1-cos(x22)-cos(x44)+cos(x22)cos(x44)]} - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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)]}`

बेरीज
Advertisements

उत्तर

Consider `1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)`

= `[1 - cos(x^2/2)] - cos(x^4/4) [1 - cos(x^2/2)]`

= `[1 - cos(x^2/2)] [1 - cos(x^4/4)]`

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

= `([1 - cos^2(x^2/2)][1 - cos^2 (x^4/4)])/([1 + cos(x^2/2)][1 + cos(x^4/4)]`

= `(sin^2(x^2/2) sin^2(x^4/4))/([1 + cos(x^2/2)][1 + cos(x^4/4)]`

∴ the required limit

= `lim_(x -> 0) (sin^2(x^2/2)sin^2(x^4/4))/(x^12[1 + cos(x^2/2)][1 + cos(x^4/4)]`

= `lim_(x -> 0) (sin^2(x^2/2) sin^2 (x^4/4))/(x^4/4 xx x^8/16 [1 + cos(x^2/2)][1 + cos(x^4/4)]) xx 1/64`

= `1/64([lim_(x -> 0) (sin(x^2/2))/((x^2/2))]^2 xx  [lim_(x -> 0) (sin(x^4/4))/((x^4/4))]^2)/([lim_(x -> 0) {1 + cos(x^2/2)}] xx [lim_(x -> 0) {1 + cos(x^4/4)}]`

= `1/64 (1^2 xx 1^2)/((1 + cos 0)(1 + cos 0))   ...[(because x -> 0  therefore x^2/2 -> 0","  x^4/4 -> 0),("and" lim_(theta -> 0) sintheta/theta = 1)]`

= `1/64 xx 1/(2 xx 2)`

= `1/256`

shaalaa.com
Concept of Limits
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Limits - Miscellaneous Exercise 7.2 [पृष्ठ १५९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 7 Limits
Miscellaneous Exercise 7.2 | Q II. (23) | पृष्ठ १५९

संबंधित प्रश्‍न

Evaluate the following limit :

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


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


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) (x^2 - 1)` = 3


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

x 1.9 1.99 1.999 2.001 2.01 2.1
f(x) 0.25641 0.25062 0.250062 0.24993 0.24937 0.24390

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 -> 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 -> 1) sin pi x`


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`


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`


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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> oo)(1 + "k"/x)^("m"/x)`


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) (sin("a" + x) - sin("a" - x))/x`


Evaluate the following limits:

`lim_(x -> 0) (2 "arc"sinx)/(3x)`


Evaluate the following limits:

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


Evaluate the following limits:

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


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) (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 -> oo) sinx/x`


Choose the correct alternative:

`lim_(x - pi/2) (2x - pi)/cos x`


Choose the correct alternative:

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


`lim_(x -> 5) |x - 5|/(x - 5)` = ______.


`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.


If `lim_(x -> 1) (x + x^2 + x^3|+ .... + x^n - n)/(x - 1)` = 820, (n ∈ N) then the value of n is equal to ______.


The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×