Re: Hints for 1


[ Follow Ups ] [ Post Followup ] [ Calculus Message Board ] [ FAQ ]

Posted by Tara on September 12, 2002 at 13:18:53:

In Reply to: Hints for 1 posted by Subhotosh Khan on September 12, 2002 at 13:01:06:

: : : : I would appreciate help evaluating the following limit:

: : : : 1) lim x->a+ ( sqrt(x) - sqrt(a) + sqrt(x-a) ) / sqrt(x^2 - a^2)

: : : : 2) lim x->0 ( (1+x^3)^(1/3) - (1+x^5)^(1/5) ) / x^3

: : : : 3)lim x->0 ( (1+ax)^(1/m) - (1+bx)^(1/n) ) / x

: :
: : Not yet! Thank you for any help you can give me!

: *************************************
: 1)( sqrt(x) - sqrt(a) + sqrt(x-a) ) / sqrt(x^2 - a^2)

: substitute:

: x = m^2 and a = n^2

: ( sqrt(x) - sqrt(a) + sqrt(x-a) ) / sqrt(x^2 - a^2)

: =[ m - n + sqrt(m^2-n^2) )] / sqrt(m^4 - n^4)

: =[ m - n]/ sqrt(m^4 - n^4) + sqrt(m^2-n^2) / sqrt(m^4 - n^4)

: =[ m - n]/ sqrt[(m - n)(m+n)(m^2+n^2)] + sqrt(m^2-n^2) / sqrt[(m^2 - n^2)(m^2+n^2)]

: = sqrt(m - n) / sqrt[(m+n)(m^2+n^2)] + 1/ sqrt(m^2+n^2)

: Now apply limit (m = n)

: = 0 + 1/ sqrt(2n^2)

: = 1/ sqrt(2a)


Thank you so much for your response!! I really appreciate it! These three problems really had me stuck - your response has helped a lot. Thanks again.



Follow Ups:



Post a Followup

Name:
E-Mail:

Subject:

Comments:

Optional Link URL:
Link Title:
Optional Image URL:


[ Follow Ups ] [ Post Followup ] [ Calculus Message Board ] [ FAQ ]