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My answer was wrong. I started the problem off by conjugate rule, but i think i mesed up somewhere with the lone values near the radical roots. And i was plugging in 0 at the end. Maybe that was wrong.

Veteran  Apr 14, 2017
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 #1
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Find the following limit:
lim_(x->0) (sqrt(4 x + 16) - 4)/(7 x)

(sqrt(4 x + 16) - 4)/(7 x) = (sqrt(4 x + 16) - 4)/(7 x):
lim_(x->0) (sqrt(4 x + 16) - 4)/(7 x)

lim_(x->0) (sqrt(4 x + 16) - 4)/(7 x) = 1/7 (lim_(x->0) (sqrt(4 x + 16) - 4)/x):
1/7 lim_(x->0) (sqrt(4 x + 16) - 4)/x

(sqrt(4 x + 16) - 4)/x = ((sqrt(4 x + 16) - 4) (4 + sqrt(4 x + 16)))/(x (4 + sqrt(4 x + 16))) = 4/(4 + sqrt(4 x + 16)):
(lim_(x->0) 4/(4 + sqrt(4 x + 16)))/(7)

lim_(x->0) 4/(4 + sqrt(4 x + 16)) = 4/(4 + sqrt(16 + 4 0)) = 1/2:
1/71/2

1/(2 7) = 1/14:
Answer: | 1/14

Guest Apr 14, 2017
 #2
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Other guest is correct ,limit is 1/14, but use L'Hopital's Theorem.  Much easier.

Limit becomes {1/2(4x+16)^(-1/2)   X   4}/7      (chain rule )

 

= { 2(4x+16)^(-1/2)} /7        now let x tend to zero 

= {2 (1/4)}  /7     = (1/2)  / 7      =     1/14

Guest Apr 14, 2017

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