Question

(Sequences): Find the minimum value of n must be greater than for the following to be true:

|(7+1/n)^2-49| < e for e >0(Sequences): Find the minimum value of n must be greater than for the following to be true:

|(7+1/n)^2-49| < e for e >0@Mathematics

Answer 1

\[\left| (7+1/n)^{2} - 49\right|<e, e>0\]

Answer 2

Just as a note, I never did the epsilon/delta stuff in Cal 1 so I am quite lost with this :(

Answer 3

So, if I'm understanding right... you could just write this as:
\[\left| (7+\frac{1}{n})^2-49 \right|<0\]
since the thing has to be less than e and e must be bigger than 0...
and as it happens that inequality has no solutions. Unless I'm missing something.

Answer 4

whats the original question in complete form

Answer 5

dont give us drips and drabs of the question, type the full complete one ;)

Answer 6

Assuming n-> infinity, which of the following conditions represents the minimum value n must be greater than in order to satisfy: |(7+.....)|<e for any e>o

Answer 7

a) n>\[(7+\sqrt{49+e})/e\]
b) n> \[(7-sqrt{49+e})/e\]
c) n> \[(7+sqrt{49-e})/e\]
d) n> \[(7-sqrt{49-e})/e\]

Answer 8

ok the sqrt is supposed to be over the 49-/+e all the time..

Answer 9

and e here is epsilon...

Answer 10

you want to show that lim ( 7 + 1/n) ^2 = 49

Answer 11

as n-> oo

Answer 12

I dunno none of my answers match... I got this: n >
7 - sqrt(e+49)
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7*sqrt(e+49)