If $-5\leq a \leq -1$ and $1 \leq b \leq 3$, what is the least possible value of $\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right) $? Express your answer as a common fraction.
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\(\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ =\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\\ =\frac{1}{b^2}-\frac{1}{a^2}\\ smallest - biggest=smallest\\ \text{so I need the biggest }b^2 \text { and the smallest }a^2\\ biggest \;\;b^2=9\qquad smallest\;\;a^2=1\\ \text{smallest value = }\frac{1}{9}-1 = -\frac{8}{9}\\ \)
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