हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Find the left and right limits of f(x) = x2-4(x2+4x+4)(x+3) at x = – 2

Advertisements
Advertisements

प्रश्न

Find the left and right limits of f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 2

योग
Advertisements

उत्तर

f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 2

f(x) = `((x + 2)(x - 2))/((x + 2)^2 (x + 3))`

f(x) = `(x - 2)/((x +2)(x +3))`

o find the let imit of f(x) at x = – 2

Put x = – 2 – h

Where h > 0

When x →  – 2

We have h → 0

`lim_(x -> - 2^-) f(x) =  lim_("h" -> 0) ((-2 - "h")- 2)/((-2  "h" + 2)(- 2 - "h" + 3)`

= `lim_("h" -> 0) (-4 - "h")/((- "h")(1 - "h"))`

=`lim_("h" -> 0) 1/"h" ((4 + "h")/(1 - "h"))`

= `1/0 ((4 + 0)/(1 - 0))`

= `oo`

`lim_(x -> - 2^-) f(x) = oo`

o find the right limit of f(x) at x = – 2

Put x = – 2 + h

Where h > 0

When x →  – 2

We have h → 0

`lim_(x -> - 2) f(x) =  lim_("h" -> 0) ((-2 + "h") - 2)/((-2 + "h" + 2)(-2 + "h" + 3))`

= `lim_("h" -> 0) (-4 + "h")/("h"(1 + "h"))`

= `lim_("h" -> 0) 1/"h"(("h" - 4)/(1 +"h"))`

= `1/0 ((0 - 4)/(1 + 0))`

= `- oo`

`lim_(x -> - 2^-) f(x) = - oo`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.3 [पृष्ठ १११]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.3 | Q 1. (a) | पृष्ठ १११

संबंधित प्रश्न

Evaluate the following limit:

`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`


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 : 

If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.


Evaluate the following :

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


In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`

x – 0.1  – 0.01 – 0.001 0.001 0.01 0.1
f(x) 0.2911 0.2891 0.2886 0.2886 0.2885 0.28631

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_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Choose the correct alternative:

If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is


Choose the correct alternative:

`lim_(x -> 0) ("e"^(sin x) - 1)/x` =


`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×