हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा 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")`

योग

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
Concept of Limits
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 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
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.2 | Q 15 | पृष्ठ १०३
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×