हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Evaluate the following limits: limx→0x2+1-1x2+16-4

Advertisements
Advertisements

प्रश्न

Evaluate the following limits:

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

योग
Advertisements

उत्तर

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

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

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

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

= `(sqrt(0^2 + 16) + 4)/(sqrt(0^2 + 1) + 1)`

= `(4 + 4)/(1 + 1)`

`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4) = 8/2` = 4

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.2 [पृष्ठ १०३]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.2 | Q 8 | पृष्ठ १०३

संबंधित प्रश्न

Evaluate the following limit :

`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 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 -> 2) (x^2 - 1)` = 3


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


Evaluate the following :

`lim_(x -> 0)[x/(|x| + x^2)]`


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)]}`


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`


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(– 2) = 0

f(2) = 0

`lim_(x -> 2) f(x)` = 0

`lim_(x -> 2) f(x)` does not exist.


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


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


Evaluate the following limits:

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


Evaluate the following limits:

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


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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (tan 2x)/x`


Choose the correct alternative:

If `f(x) = x(- 1)^([1/x])`, x ≤ 0, then the value of `lim_(x -> 0) f(x)` is equal to


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 = ______.


`lim_(x -> 0) (sin 4x + sin 2x)/(sin5x - sin3x)` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×