मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Evaluate the following limits: ababaablimx→ax-b-a-bx2-a2(a>b) - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following limits:

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

बेरीज
Advertisements

उत्तर

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

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

= `lim_(x -> "a") ((x - "b") - ("a"- "b"))/((x^2 - "a"^2) [sqrt(x - "b") + sqrt("a" - "b")]`

= `lim_(x -> "a") (x - "b" - "a" + "b")/((x - "a")(x + "a") [sqrt(x - "b") + sqrt("a" - "b")]`

= `lim_(x -> "a") (x - "a")/((x - "a")(x + "a") [sqrt(x - "b") + sqrt("a" - "b")]`

= `lim_(x -> "a") 1/((x + "a")[sqrt("x" - "b") + sqrt('a" -"b")]`

= `1/(("a" + "a")[sqrt("a" - "b") + sqrt("a" - "b")]`

= `1/(2"a" xx 2sqrt("a" - "b")`

= `1/(4"a"sqrt("a" - "b")`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 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
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.2 | Q 15 | पृष्ठ १०३

संबंधित प्रश्‍न

Evaluate the following limit:

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


Evaluate the following limit :

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


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 problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`

x – 3.1  – 3.01 – 3.00 – 2.999 – 2.99 – 2.9
f(x) – 0.24845 – 0.24984 – 0.24998 – 0.25001 – 0.25015 – 0.25158

Evaluate the following limits:

`lim_(x ->) (x^"m" - 1)/(x^"n" - 1)`, m and n are integers


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`


Evaluate the following limits:

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


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


Evaluate the following limits:

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


Evaluate the following limits:

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


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) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> ) (sinx(1 - cosx))/x^3`


Evaluate the following limits:

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


Choose the correct alternative:

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×