Question

lim
x→−7 of

7 − |x| / 7 + x

Answer 1

Do you mean \(\dfrac{7-|x|}{7+x}\), because that is very much NOT what you have written. You have \(7 - \dfrac{|x|}{7} + x\). Remember your Order of Operations.

Answer 2

yes i meant what you think i meant but i had a hard time typing sorry

Answer 3

limit((7-abs(x))/(7+x), x = -7) = 1

Answer 4

tkhunny can you please help me

Answer 5

Kind of a fun little idea on this one. We have the usual...

If x >= 0, then |x| = x
If x < 0, then |x| = -x

However, Evaluating the limit, eventually we get nowhere near x = 0 and we can just ignore the part about x >= 0.

Considering, then, ONLY x < 0, we have \(\lim_{x \rightarrow -7}\dfrac{7 - |x|}{7+x} = \lim_{x \rightarrow -7}\dfrac{7 - (-x)}{7+x}\) and the whole thing is a lot simpler.

Answer 6

wait... i thought that the absolute value of ANY number positive or negative is always positive

Answer 7

the absolutely value of -2 is 2

Answer 8

if what you are saying is correct then the evaluated number would be 0 ???

Answer 9

|-2| = -(-2) = +2

If x < 0, |x| = -x

Answer 10

im still confused

Answer 11

If x > 0, |x| = x
Example
|5| = 5

If x < 0, |x| = -x
Example
|-3| = -(-3) = +3

You have to get his idea functionally in your head.

Essentially, the absolute value does nothing to positive numbers.
The absolute value changes the sign on negative numbers.

Answer 12

ok i understand.. but i still dont understand my initial problem

Answer 13

\(\dfrac{7-(-x)}{7+x} = \dfrac{7+x}{7+x} = 1,\;except\;where\;x = -7.\) It's not continuous, but the limit exists.