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when v->4+ means v is very very near to 4, but greater than 4. agreed?
so, 4-v will be positive or negative?
thats right :) and since the denominator is having a modulus, it will be positive only. so our final answer will be negative!
I understand by writing things out..
i meant |4-v| is always positive and 4-v is negative
So how would I solve it?
4-v from numerator and denominator would cancel out, so the answer would be either +1 or -1 the numerator is negative, and denominator is positive, so the overall answer is negative, that is -1.
Are you sure it's that simple?
|dw:1439231746924:dw|Is that how the work would look like?
ok, lets make some steps, since v is very near to 4, but greater than 4, lets take v = 4 + \(\delta \) when \(v \to 4^+\), \(\delta \to 0\) |dw:1439231931131:dw| try to simplify that ^^
the numerator simplifies to \(-\delta\) and the denominator simplifies to \(|-\delta | = \delta\) isn't it ?
I see it now! thanks for the clear work