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Simplifying limits with multiple square roots in fractions?

+1 vote
Hi, from my working I reached the equation line

lim(h→0)⁡ = (2√(t+h)-2√t) / h
asked Dec 26, 2012 in PRECALCULUS by johnkelly Apprentice

1 Answer

+2 votes

  lt       (2sqrt (t+h)-2sqrt (t)) / h

h→0     

= (2sqrt (t+0)-2sqrt (t))/0

= (2sqrt (t)-2sqrt (t))/0

= 0/0

Since the numarator and denominator are zeros.the function is undefined.

By using L-hospital rule

  lt       (2sqrt (t+h)-2sqrt (t))/h

h→0 

=  lt       (2t 1/2sqrt (t+h)  -2sqrt (t)) / 1

   h→0

=  lt       (t/sqrt (t+h)  -2sqrt (t)) / 1

   h→0

=  (t/sqrt (t+0)  - 2sqrt (t)) / 1

=  (t/sqrt (t) - 2sqrt (t)) / 1

= (sqrt (t) - 2sqrt (t)) / 1

= (sqrt (t) - 2sqrt (t))

= - sqrt (t)

  lt       (2sqrt (t+h)-2sqrt (t)) / h  =  - sqrt (t)

h→0

I hope it helps. if you like this answer please select as best answer, thanks.

answered Dec 26, 2012 by friend Mentor

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