हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Evaluate the following limits: limx→0sinαxsinβx - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following limits:

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

योग
Advertisements

उत्तर

We know `lim_(x -> 0) sinx/x` = 1

`lim_(x -> 0) (sin alpha x)/(sin betax) =  lim_(x -> 0) (sin alphax)/(1/alpha (alphax)) xx (1/beta (betax))/(sin betax)`

= `alpha/beta lim_(x -> 0) (sin(alphax))/((alphax)) xx (betax)/(sin(betax))` 

= `alpha/beta lim_(alphax -> 0) (sin(alphax))/(alphax) xx lim_(betax -> 0) (betax)/(sin(betax))`

= `alpha/beta lim_(alphax -> 0) (sin(alphax))/(alphax) xx 1/(lim_(betax -> 0) (sin("betax))/(betax))`

= `alpha/beta xx 1 xx 1/1`

`lim_(x -> 0) (sinalphax)/(sinbetax) = alpha/beta`

shaalaa.com
Concept of Limits
  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 7 | पृष्ठ ११८

संबंधित प्रश्न

Evaluate the following limit:

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


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 -> -3) (3x + 2)` = – 7


Evaluate the following :

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


Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.

f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`


Sketch the graph of a function f that satisfies the given value:

f(0) is undefined

`lim_(x -> 0) f(x)` = 4

f(2) = 6

`lim_(x -> 2) f(x)` = 3


Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",",  "for"  x ≠ 1),(0",",  "for"  x = 1):}`


Find the left and right limits of f(x) = tan x at x = `pi/2`


Evaluate the following limits:

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


Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1


Evaluate the following limits:

`lim_(x -> oo)(1 + 1/x)^(7x)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`


Evaluate the following limits:

`lim_(x -> pi) (1 + sinx)^(2"cosec"x)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×