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Question

How do you find the limit of $$\left ( \frac{1}{1 + h} \right ) - \frac{1}{h}$$ as h approaches $$0$$?


Solution

Answer is,
$$\lim_{h\rightarrow 0^-} \left ( \frac{1}{1 + h} \right ) - \frac{1}{h} = - \infty $$
Explanation:
$$\lim_{h\rightarrow 0^-} \left ( \frac{1}{1 + h} \right ) - \frac{1}{h}$$
$$\lim _{ h\rightarrow 0^{ - } } \frac { h-\left( 1+h \right)  }{ h(1+h) } $$
$$\lim_{h\rightarrow 0^-} \frac{-1}{h (1 + h)}$$$$=-\frac { 1 }{ 0 } =-\infty $$

$$\therefore \lim_{h\rightarrow 0^-} \left ( \frac{1}{1 + h} \right ) - \frac{1}{h} = - \infty $$

Mathematics

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