# What is the limit of  sinx /(x^2 - 4x) as x approaches 0?

##### 2 Answers
Sep 1, 2015

${\lim}_{x \to 0} \sin \frac{x}{{x}^{2} - 4 x} = - \frac{1}{4}$

#### Explanation:

You know that $\sin 0 = 0$, so you cannot evaluate this limit by replacing $x$ with zero, since that would get you

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{\lim}_{x \to 0} \frac{\sin 0}{{0}^{2} - 4 \cdot 0} = \frac{0}{0}}}} \to$ indeterminate form

This means that you're going to have to use L'Hopital's Rule to find this limit. According to L'Hopital's rule, if you have two functions $f \left(x\right)$ and $g \left(x\right)$ which are differentiable on an interval $\left(a , b\right)$, and if

${\lim}_{x \to c} \frac{{f}^{'} \left(x\right)}{{g}^{'} \left(x\right)} \text{ }$, with ${g}^{'} \left(x\right) \ne 0$ and $c \in \left(a , b\right)$

exists, then you have

color(blue)(lim_(x->0)(f(x))/(g(x)) = lim_(x->0)(f^'(x))/(g^'(x))

In your case, you have

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x \text{ }$ and $\text{ } \frac{d}{\mathrm{dx}} \left({x}^{2} - 4 x\right) = 2 x - 4$

which means that

${\lim}_{x \to 0} \sin \frac{x}{{x}^{2} - 4 x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(\sin x\right)}{\frac{d}{\mathrm{dx}} \left({x}^{2} - 4 x\right)} = \cos \frac{x}{2 x - 4}$

Now you can evaluate this limit for $x \to 0$ by replacing $x$ with zero

${\lim}_{x \to 0} \frac{\cos 0}{2 \cdot 0 - 4} = \frac{1}{\left(- 4\right)} = - \frac{1}{4}$

Therefore

${\lim}_{x \to 0} \sin \frac{x}{{x}^{2} - 4 x} = \textcolor{g r e e n}{- \frac{1}{4}}$

Sep 1, 2015

${\lim}_{x \rightarrow 0} \sin \frac{x}{{x}^{2} - 4 x} = - \frac{1}{4}$

#### Explanation:

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

$= {\lim}_{x \rightarrow 0} \sin \frac{x}{x} \frac{1}{\left(x - 4\right)}$

$= {\lim}_{x \rightarrow 0} \sin \frac{x}{x} {\lim}_{x \rightarrow 0} \frac{1}{\left(x - 4\right)}$ $\text{ }$if both limits exist

$= \left(1\right) \left(- \frac{1}{4}\right)$

$= - \frac{1}{4}$