मराठी
तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Evaluate the following limits: aabblimx→0x2+a2-ax2+b2-b - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following limits:

`lim_(x -> 0) (sqrt(x^2 + "a"^2) - "a")/(sqrt(x^2 + "b"^2) - "b")`

बेरीज
Advertisements

उत्तर

We know `lim_(x -> "a") (x^"n" - "a"^"a")/(x -"a") = "na"^("n" - 1)`

`lim_(x -> 0)((sqrt((x^2 + "a"^2)) - "a")/(sqrt((x^2 + "b"^2)) - "b")) =  lim_(x -> 0)(((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2)))`

= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/x^2 xx x^2/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))`

= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "a"^2) - "a"^2) xx ((x^2 + "b"^2) - "b"^2)/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))`

= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "a"^2) - "a"^2) xx lim_(x -> 0) 1/(((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))/((x^2 + "b"^2) - "b"^2)`

Put x2 + a2 = y

Put x2 + b2 = z

When x = 0

⇒ y = a2

When x = 0

⇒ z = b2 

= `lim_(y -> "a"^2)((y^(1/2) -  ("a"^2)^(1/))/(y - "a"^2)) xx 1/(lim_(x -> "b"^2)(("z"^(1/2) - ("b"^2)^(1/2))/("z" - "b"^2))`

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

= `(("a"^2)^(- 1/2))/(("b"^2)^(- 1/2))`

= `("a"^(- 1))/("b"^(- 1))`

= `"b"/"a"`

`lim_(x -> 0)((sqrt((x^2 + "a"^2)) - "a")/(sqrt((x^2 + "b"^2)) - "b")) = "b"/"a"`

shaalaa.com
Concept of Limits
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.4 [पृष्ठ ११८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.4 | Q 11 | पृष्ठ ११८

संबंधित प्रश्‍न

Evaluate the following limit:

`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`


Evaluate the following limit :

`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`


Evaluate the following limit :

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


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 -> 0) sec x`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (sqrt(1 + x) - 1)/x`


Evaluate the following limits:

`lim_(x -> 5) (sqrt(x - 1) - 2)/(x - 5)`


Evaluate the following limits:

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


Show that `lim_("n" -> oo) (1 + 2 + 3 + ... + "n")/(3"n"^2 + 7n" + 2) = 1/6`


Show that  `lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = 9/25`


Evaluate the following limits:

`lim_(x -> oo)(1 + 1/x)^(7x)`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`


Choose the correct alternative:

`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =


`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.


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×