हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Sketch the graph of f, then identify the values of x0 for which limx→x0 f(x) exists. f(x) = ,,,{x2,x≤28-2x,2<x<44,x≥4 - Mathematics

Advertisements
Advertisements

प्रश्न

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):}`

सारिणी
आलेख

उत्तर

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

x 0 1 2 3 3.5 4 5 6
f(x) x2 x2 x2 8 – 2x 8 – 2x 4 4 4
f(x) 0 1 4 2 1 4 4 4


At x = 4, the curve does not exist.

Hence, except at x0 = 4, the limit of f(x) exists.

shaalaa.com
Concept of Limits
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९७]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 16 | पृष्ठ ९७

संबंधित प्रश्न

Evaluate the following limit:

`lim_(z -> -3) [sqrt("z" + 6)/"z"]`


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


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):}`


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

f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`


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


Evaluate the following limits:

`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`


Evaluate the following limits:

`lim_(x -> 0) (tan 2x)/(sin 5x)`


Evaluate the following limits:

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


Choose the correct alternative:

`lim_(x -> oo) sinx/x`


Choose the correct alternative:

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


Choose the correct alternative:

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


`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.


The value of `lim_(x rightarrow 0) (sqrt((1 + x^2)) - sqrt(1 - x^2))/x^2` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×