# How do solve without using derivative?

## $\lim x \rightarrow 0 \frac{\sin \left({x}^{3} + {x}^{2} - x\right) + \sin x}{x}$

Dec 16, 2016

${\lim}_{x \to 0} \frac{\sin \left({x}^{3} + {x}^{2} - x\right) + \sin x}{x} = 0$

#### Explanation:

You can resolve this limit without using derivatives by re-conducing it to the known limit:

${\lim}_{x \to 0} \sin \frac{x}{x} = 1$

We can proceed in this way:

$\frac{\sin \left({x}^{3} + {x}^{2} - x\right) + \sin x}{x} = \frac{\sin \left({x}^{3} + {x}^{2} - x\right)}{x} + \frac{\sin x}{x} = \frac{\sin \left({x}^{3} + {x}^{2} - x\right)}{{x}^{3} + {x}^{2} - x} \cdot \frac{{x}^{3} + {x}^{2} - x}{x} + \frac{\sin x}{x} = \frac{\sin \left({x}^{3} + {x}^{2} - x\right)}{{x}^{3} + {x}^{2} - x} \cdot \left({x}^{2} + x - 1\right) + \frac{\sin x}{x}$

Thus:
${\lim}_{x \to 0} \frac{\sin \left({x}^{3} + {x}^{2} - x\right) + \sin x}{x} = {\lim}_{x \to 0} \frac{\sin \left({x}^{3} + {x}^{2} - x\right)}{{x}^{3} + {x}^{2} - x} \cdot \left({x}^{2} + x - 1\right) + \frac{\sin x}{x} = 1 \cdot \left(- 1\right) + 1 = 0$