Show that a sequence {sn} coverages to a limit l if and only if the sequence {sn-l} coverages to zero.

Step-by-step explanation:

To prove this we can use the definition of a sequence converging to its limit, in terms of epsilon:

The sequence converges to

if and only if

for every there exists such that

if and only if

for every there exists such that

if and only if

the sequence converges to 0.

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it