# lim_(x to 0) sin^5(x)/ (x^2+x) sin(x^4) how should i solve it?

## im supposed to use the ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$ or ${\lim}_{u \left(x\right) \to 0} \sin \frac{u \left(x\right)}{u \left(x\right)} = 1$ formulas

Jan 28, 2018

$0$

#### Explanation:

${\lim}_{x \to 0} {\sin}^{5} \frac{x}{{x}^{2} + x} \sin \left({x}^{4}\right)$

Take the ${\sin}^{5} \left(x\right)$ and split it up like so:

$= {\lim}_{x \to 0} \sin \frac{x}{{x}^{2} + x} {\sin}^{4} \left(x\right) \sin \left({x}^{4}\right)$

$= {\lim}_{x \to 0} \sin \frac{x}{{x}^{2} + x} {\lim}_{x \to 0} {\sin}^{4} \left(x\right) \sin \left({x}^{4}\right)$

So, evaluating these limits separately :

${\lim}_{x \to 0} {\sin}^{4} \left(x\right) \sin \left({x}^{4}\right) = 0 \times 0$

and

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

For the denominator, the ${x}^{2}$ term will go to $0$ much faster than the $x$ term so it will vanish leaving us with:

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

$= {\lim}_{x \to 0} \frac{\sin \left(x\right)}{x} = 1$

Hence:

${\lim}_{x \to 0} {\sin}^{5} \frac{x}{{x}^{2} + x} \sin \left({x}^{4}\right)$

$= {\lim}_{x \to 0} \sin \frac{x}{{x}^{2} + x} {\lim}_{x \to 0} {\sin}^{4} \left(x\right) \sin \left({x}^{4}\right)$

$1 \times 0 \times 0 = 0$