Calc 101 Question on simplifying a fraction

Question

$$\lim_{h \to 0} \left(\frac 1h -\dfrac{1}{h^2+h} \right).$$ What do I do about the denominators?

Answer 1

To add or subtract fractions, you put them over a common denominator. Can you do that here?

Answer 2

$$\dfrac 1h -\dfrac{1}{h^2+h} = \dfrac 1h - \dfrac{1}{h(h + 1)}$$ So a common denominator is given by $h(h+1) = h^2 + h$.

Express the first fraction as an equivalent fraction with this common denominator $$\dfrac 1h = \dfrac{h + 1}{h(h+1)}$$ then take the numerator of this fraction and subtract that of the second fraction, all over the the common denominator.

$$\dfrac 1h -\dfrac{1}{h^2+h} = \dfrac 1h - \dfrac{1}{h(h + 1)}=\dfrac{(h+1)-1}{h(h+1)} = \dfrac{h}{h(h+1)} = \dfrac{1}{h+1}$$