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Posted by Subhotosh Khan on September 12, 2002 at 13:01:06:

In Reply to: Re: Have you learnt L'Hospital's rule? (n/t) - No we haven't done that yet!!!! posted by Tara on September 12, 2002 at 11:49:16:

: : : 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)

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