हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why? limx→1f(x) where ,,f(x)={x2+2,x≠11,x=1

Advertisements
Advertisements

प्रश्न

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):}`

आलेख
Advertisements

उत्तर

`f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`

To find `lim_(x -> 1) f(x)`

From the graph the value of the function is y = f(1) = 3

∴ `lim_(x -> 1) f(x)` = 3

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९६]

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 10 | पृष्ठ ९६

संबंधित प्रश्न

Evaluate the following limit :

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`


Evaluate the following limit :

`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 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 -> -3) (3x + 2)` = – 7


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 exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

`lim_(x -> 3) (4 - 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 -> 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 -> x/2) tan x`


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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`


Evaluate the following limits:

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


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


Choose the correct alternative:

`lim_(x -> 0) (x"e"^x - sin x)/x` is


If `lim_(x->1)(x^5-1)/(x-1)=lim_(x->k)(x^4-k^4)/(x^3-k^3),` then k = ______.


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×